9 Low-Q harmonic oscillators

Q 9.1

(a)

An answer:
This 3 stages consist of a unity gain voltage buffer that each have their own C-R high-pass section:

The signal transfer of each voltage buffer preceded by a CR-high pass is:

$$\begin{array}{llll}\hfill H(\mathit{j\omega })=\frac{R}{R+\frac{1}{\mathit{j\omega C}}}& =\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}& \hfill & \end{array}$$

(b)

An answer:

With the previous answer, the signal transfer function for the 3 cascaded sections is:

$$\begin{array}{llll}\hfill \frac{v_z}{v_{\mathit{out}}}& ={\left(\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\right)}^3& \hfill & \\ \hfill & =\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{(1+\mathit{j\omega RC})\cdot (1+2\mathit{j\omega RC}-{\omega }^2{(\mathit{RC})}^2)}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{1+3\mathit{j\omega RC}-3{\omega }^3{(\mathit{RC})}^2-j{\omega }^3{(\mathit{RC})}^3}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{\{1-3{\omega }^2{(\mathit{RC})}^2\}+j(3\mathit{\omega RC}-{\omega }^3{(\mathit{RC})}^3)}& \hfill & \end{array}$$

(c)

An answer:

(d)

An answer:
Node $x$. From the Bode pot for the 3 cascaded stages it follows that for oscillation at a finite non-zero frequency, the gain of the leftmost (amplifying) stage must be negative. There is just one (radian) frequency at which $\mathit{\angle H}(\mathit{j\omega })=\pm 180^{\circ }$; hence there is no need to check whether other frequencies would be more favourable for oscillation.

(e)

An answer:

$$\begin{array}{llll}\hfill 1-3{\omega }_{\mathit{osc}}^2{(\mathit{RC})}^2& =0& \hfill & \\ \hfill \to {\omega }_{\mathit{osc}}^2& =\frac{1}{3{(\mathit{RC})}^2}& \hfill & \\ \hfill \to {\omega }_{\mathit{osc}}& =\frac{1}{\sqrt{3}\mathit{RC}}& \hfill & \end{array}$$

(f)

An answer:

$$\begin{array}{llll}\hfill \frac{v_z}{v_{\mathit{out}}}\left({\omega }_{\mathit{osc}}=\frac{1}{\sqrt{3}\mathit{RC}}\right)& =\frac{-j{\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)}^3{(\mathit{RC})}^3}{j\left(3\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)\mathit{RC}-{\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)}^3{(\mathit{RC})}^3)\right)}& \hfill & \\ \hfill & =\frac{-{(\frac{1}{\sqrt{3}})}^3}{\{\frac{3}{\sqrt{3}}-{(\frac{1}{\sqrt{3}})}^3\}}& \hfill & \\ \hfill & =-\frac{\frac{1}{3\sqrt{3}}}{\frac{3}{\sqrt{3}}-\frac{1}{3\sqrt{3}}}& \hfill & \\ \hfill & =-\frac{1}{8}& \hfill & \end{array}$$

The voltage gain of the leftmost gain stage must equal -8. For an opamp in negative feedback configuration then $\beta =8$.

Q 9.2

(a)

An answer:

$$\begin{array}{lll}\hfill H_{\beta }(\mathit{j\omega })=\frac{v_{+}-v_{-}}{v_{\mathit{out}}}=\frac{\mathit{j\omega }\left(R_{2}C-\frac{L}{R_1}\right)}{1+\mathit{j\omega }\left(R_{2}C+\frac{L}{R_1}\right)+j^2{\omega }^2\frac{R_2}{R_1}\mathit{CL}}& & \hfill \end{array}$$

(b)

An answer:

$$\begin{array}{lll}\hfill {\omega }_{\mathit{osc}}=\sqrt{\frac{R_1}{R_2\mathit{LC}}}& & \hfill \end{array}$$

(c)

An answer:

$$\begin{array}{lll}\hfill A=\frac{R_1{R}_{2}C+L}{R_1{R}_{2}C-L}& & \hfill \end{array}$$

Q 9.3

(a)

An answer:
A is negative, so the combined phase shift of the 5 $\mathit{RC}$ filters should be -180 degrees. This leads to $\mathit{\angle stage}=-36^{\circ }$

(b)

An answer:

$$\begin{array}{lll}\hfill {\omega }_{\mathit{osc}}=\frac{\mathit{tan}(36^{\circ })}{\mathit{RC}}& & \hfill \end{array}$$

(c)

An answer:

$$\begin{array}{lll}\hfill A=-\sqrt{1+\mathit{tan}{(36^{\circ })}^2}& & \hfill \end{array}$$

Q 9.4

(a)

An answer:

$$\begin{array}{llll}\hfill H_{\beta }(\mathit{j\omega })=\frac{v_{\mathit{in}}}{v_{\mathit{out}}}& ={\left(\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\right)}^3& \hfill & \end{array}$$

(b)

An answer:

For the Bode plot, we defined ${\omega }_0=\frac{1}{\mathit{RC}}$.

(c)

An answer:
Using symmetry is the easiest and leads to:

$$\begin{array}{rcll}& {\omega }_{\mathit{osc}}& =\frac{1}{\mathit{RC}}\mathit{tan}(30^{\circ })& \\ & & =\frac{1}{\sqrt{3}\mathit{RC}}& \end{array}$$

(d)

An answer:
A = -8

(e)

An answer:
The polar plot of $\mathit{A\beta }$ is shown in blue for positive $\omega$; the light blue curve is for negative frequencies and is required to make it a Nyquist plot.

Q 9.5

(a)

An answer:
$A=-\frac{g_{m}R}{1+\mathit{j\omega RC}}$.

(b)

An answer:

$$\begin{array}{rcll}& A_{\mathit{OL}}& =A^3(3\mathit{identical}\mathit{stages})& \\ & & ={\left(\frac{-g_m\cdot R}{1+\mathit{j\omega RC}}\right)}^3& \\ & & g_m=\frac{q}{\mathit{kT}}\cdot I_C\approx 40I_C& \\ & & I_C=\frac{V_{\mathit{CC}}-0.6}{R}& \\ & A_{\mathit{OL}}& =-{\left(\frac{40V_{\mathit{CC}}-24}{1+\mathit{j\omega RC}}\right)}^3& \end{array}$$

(c)

An answer:

$$\begin{array}{rcll}& \mathit{Arg}(A_{\mathit{OL}}(f))& =\pi +3\mathit{arg}\left(\frac{1}{1+\mathit{j\omega RC}}\right)& \\ & & =\pi -3\mathit{arg}(1+\mathit{j\omega RC})\equiv 0& \\ & & \to & \\ & {\omega }_{\mathit{osc}}& =\frac{\sqrt{3}}{\mathit{RC}}& \\ & f_{\mathit{osc}}& =\frac{{\omega }_{\mathit{osc}}}{2\pi }=\frac{\sqrt{3}}{2\mathit{\pi RC}}& \end{array}$$

(d)

An answer:

$$\begin{array}{rcll}& A_{\mathit{OL}}({\omega }_{\mathit{osc}})& =-{\left(\frac{40V_{\mathit{CC}}-24}{1+j\sqrt{3}}\right)}^3\equiv 1& \\ & & =-\frac{{(40V_{\mathit{CC}}-24)}^3}{-8}=1& \\ & \to & & \\ & V_{\mathit{CC}}=& \frac{26}{40}=0.65V& \end{array}$$

(e)

An answer:
If $V_{\mathit{CC}}$ is lower, $A_{\mathit{OL}}<1$: the circuit is stable, there is no oscillation

If $V_{\mathit{CC}}$ is higher, $A_{\mathit{OL}}>1$: the circuit becomes unstable

(f)

An answer:
4 stages: amplifiers give ${(-g_m\cdot R)}^4>0$ of gain at DC. For oscillation, the RC-networks have provide a total phase shift of $0^{\circ }$ (or $k\cdot 360^{\circ }$) phase shift. Each RC-network individually then has to provide $0^{\circ },\pm 90^{\circ },\pm 180^{\circ },...$.

The only options are then $0^{\circ }$ (at DC) which is not regarded as oscillation AND $-90^{\circ }$ (at $\omega \to \infty$) which is also not regarded as harmonic oscillation.

5 stages: this situation is very similar to 3 stages ${(-g_m\cdot R)}^5<0\to \pm 180^{\circ }$ shift. Hence, for oscillation the 5 RC-networks have to give a total of $\pm 180^{\circ }\pm k\cdot 360^{\circ }$ phase shift.

The total phase shift of the 5 RC-networks is in the range $[0^{\circ },-450^{\circ }>$. The only possibility is then to oscillate at the frequency where the phase shift per RC-network is $\frac{-180^{\circ }}{5}=-36^{\circ }$ which is possible for a non-zero finite $\omega$: 5 stages can yield harmonic oscillation.

6 stages: As for 4 stages, amplifiers give $>0$ gain at DC. The only options are then $0^{\circ }$ (=at DC) which is not regarded as oscillation AND $-60^{\circ }$. This latter frequency is finite and non-zero BUT the loop-gain of the DC-solution is larger and therefore wins. This 6-stage version cannot oscillate harmonically.

Q 9.6

(a)

An answer:

$$\begin{array}{rcll}& H_{\mathit{loop}}(\mathit{j\omega })& =A\cdot H_1(\mathit{j\omega })\cdot H_2(\mathit{j\omega })& \\ & & H_1(\mathit{j\omega })=\frac{1∕\mathit{j\omega }{C}_1}{R_1+1∕\mathit{j\omega }{C}_1}=\frac{1}{1+\mathit{j\omega }{C}_1{R}_1}& \\ & & H_2(\mathit{j\omega })=\frac{R_2}{R_2+1∕\mathit{j\omega }{C}_1}=\frac{\mathit{j\omega }{R}_2{C}_2}{1+\mathit{j\omega }{C}_2{R}_2}& \\ & H_{\mathit{loop}}(\mathit{j\omega })& =A\cdot \frac{1}{1+\mathit{j\omega }{C}_1{R}_1}\cdot \frac{\mathit{j\omega }{R}_2{C}_2}{1+\mathit{j\omega }{C}_2{R}_2}& \\ & & =A\cdot \frac{\mathit{j\omega }{R}_2{C}_2}{1+\mathit{j\omega }{R}_2{C}_2+\mathit{j\omega }{R}_1{C}_1-{\omega }^2{R}_1{C}_1{R}_2{C}_2}& \\ & & =A\cdot \frac{1}{\frac{1}{\mathit{j\omega }{R}_2{C}_2}+1+\frac{R_1{C}_1}{R_2{C}_2}+\mathit{j\omega }{R}_1{C}_1}& \end{array}$$

yielding:

$$\begin{array}{rcll}& {\omega }_{\mathit{osc}}& =\frac{1}{\sqrt{R_1{C}_1{R}_2{C}_2}}& \end{array}$$

(b)

An answer:

$$\begin{array}{rcll}& H_{\mathit{loop}}({\omega }_{\mathit{osc}})& =A\cdot \frac{1}{1+\frac{R_1{C}_1}{R_2{C}_2}}& \\ & A& =1+\frac{R_1{C}_1}{R_2{C}_2}& \end{array}$$

Q 9.7

(a)

An answer:

$$\begin{array}{rcll}& H_{\mathit{loop}}(\mathit{j\omega })& =A\cdot H_1(\mathit{j\omega })\cdot H_2(\mathit{j\omega })& \\ & & H_1(\mathit{j\omega })=\frac{R_1}{R_1+\mathit{j\omega L}}=\frac{1}{1+\mathit{j\omega }\frac{L}{R_1}}& \\ & & H_2(\mathit{j\omega })=\frac{R_2}{R_2+1∕\mathit{j\omega C}}=\frac{\mathit{j\omega }{R}_{2}C}{1+\mathit{j\omega C}{R}_2}& \\ & H_{\mathit{loop}}(\mathit{j\omega })& =A\cdot \frac{1}{1+\mathit{j\omega }\frac{L}{R_1}}\cdot \frac{\mathit{j\omega }{R}_{2}C}{1+\mathit{j\omega C}{R}_2}& \\ & & =A\cdot \frac{\mathit{j\omega }{R}_{2}C}{1+\mathit{j\omega }(\frac{L}{R_1}+R_{2}C)-{\omega }^2\frac{R_2}{R_1}\mathit{LC}}& \end{array}$$

$$\begin{array}{rcll}& {\omega }_{\mathit{osc}}& =\frac{1}{\sqrt{\frac{R_2}{R_1}\mathit{LC}}}& \end{array}$$

(b)

An answer:

$$\begin{array}{llll}\hfill & A=1+\frac{L}{R_1{R}_{2}C}& \hfill & \end{array}$$

Q 9.8

(a)

(b)

An answer:
For this circuit, the opamp is ideal and then the signal transfer per stage is $H(\mathit{j\omega })=-Z_2∕Z_1$ where $Z_1$ is between the inverting input and the output node. This yields e.g.: $$\begin{array}{rcll}& H_1(\mathit{j\omega })& =-\frac{R}{R+\mathit{j\omega L}}=-\frac{1}{1+\mathit{j\omega L}∕R}=...& \\ & H_2\mathit{j\omega })& =-\frac{R}{R+1∕\mathit{j\omega C}}=-\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}=...& \\ & H_3(\mathit{j\omega })& =-\frac{R∕∕1∕(\mathit{j\omega C})}{R}=-\frac{1}{1+\mathit{j\omega RC}}=...& \end{array}$$

The loop gain is then:

$$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot \frac{1}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{1}{1+\mathit{j\omega RC}}& \end{array}$$

(c)

An answer:
The loop gain was derived earlier. If this can be equation to 1 for a finite non-zero $\omega$ then this can omplement a harmonic oscillator. Recycling the derived equation: $$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot \frac{1}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{1}{1+\mathit{j\omega RC}}& \\ & & -A_{\mathit{buf}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega }(L∕R+2\mathit{RC})+{(\mathit{j\omega })}^2(2\mathit{LC}+R^2{C}^2)+{(\mathit{j\omega })}^3\mathit{RL}{C}^2}& \end{array}$$

This has an imaginary numerator and a complex polynomial denominator. Setting the latter to a purely imaginary (complex) number:

$$\begin{array}{rcll}& & 1+{(j{\omega }_{\mathit{osc}})}^2(2\mathit{LC}+R^2{C}^2)\equiv 0\to & \\ & & {\omega }_{\mathit{osc}}=\frac{1}{\sqrt{2\mathit{LC}+R^2{C}^2}}& \end{array}$$

Hence it is possible to get harmonic oscillation (out of the box) for this circuit for a specific scalar $A_{\mathit{buf}}$.

(d)

An answer:
Recycling the results at (c), we only need to derive $A_{\mathit{buf}}$: $$\begin{array}{rcll}& A_{\mathit{loop}}(j{\omega }_{\mathit{osc}})& =-A_{\mathit{buf}}\cdot \frac{j{\omega }_{\mathit{osc}}\mathit{RC}}{j{\omega }_{\mathit{osc}}(L∕R+2\mathit{RC})+{(j{\omega }_{\mathit{osc}})}^3\mathit{RL}{C}^2}& \\ & & =-A_{\mathit{buf}}\cdot \frac{j{\omega }_{\mathit{osc}}\mathit{RC}}{j{\omega }_{\mathit{osc}}(L∕R+2\mathit{RC})+{(j{\omega }_{\mathit{osc}})}^3\mathit{RL}{C}^2}& \\ & & =-A_{\mathit{buf}}\cdot \frac{\mathit{RC}}{(L∕R+2\mathit{RC})-{\omega }_{\mathit{osc}}^2\mathit{RL}{C}^2}\equiv 1& \\ & & & \end{array}$$

yielding a nasty equation. Full derivation:

$$\begin{array}{rcll}& A_{\mathit{buf}}& =-\frac{(L∕R+2\mathit{RC})-{\omega }_{\mathit{osc}}^2\mathit{RL}{C}^2}{\mathit{RC}}& \\ & & =-\frac{(L∕R+2\mathit{RC})-\frac{\mathit{RL}{C}^2}{2\mathit{LC}+R^2{C}^2}}{\mathit{RC}}& \\ & & & \end{array}$$

Further cleanup/simplication is NOT required.

(e)

An answer:
Not applicable.

Q 9.9

(a)

An answer:
This is an intro question - meant to get you going.

(b)

An answer:
For this circuit, the opamp is ideal and then the signal transfer per stage is $H(\mathit{j\omega })=-Z_2∕Z_1$ where $Z_1$ is between the inverting input and the output node. This yields e.g.: $$\begin{array}{rcll}& H_1(\mathit{j\omega })& =-\frac{R+\mathit{j\omega L}}{R}=-(1+\mathit{j\omega L}∕R)=...& \\ & H_2(\mathit{j\omega })& =-\frac{R+1∕(\mathit{j\omega C})}{R}=-\frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}=...& \\ & H_3\mathit{j\omega })& =-\frac{R∕∕\mathit{j\omega L}}{R}=-\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}=...& \\ & & & \end{array}$$

The loop gain is then:

$$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega L}∕R)\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \end{array}$$

(c)

An answer:
Recycling the (partial) results from (b) and simplifying: $$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega L}∕R)\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \\ & & =-A_{\mathit{buf}}\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \mathit{j\omega L}∕R& \\ & & =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega RC})\cdot L∕R^{2}C& \end{array}$$

for oscillation:

$$\begin{array}{rcll}& & =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega RC})\cdot L∕R^{2}C=1& \end{array}$$

yields $\omega =0$ which is NOT a harmonic oscillator. $\to$ (e)

(d)

(e)

An answer:
We can (sensibly) swap in just 1 way per stage. And the swap should be done in only one stage. Swapping in one stage the transfer for that stage becomes: $$\begin{array}{rcll}& H_1(\mathit{j\omega })& =-\frac{2R}{\mathit{j\omega L}}=-\frac{2}{\mathit{j\omega L}∕R}=...& \\ & H_2(\mathit{j\omega })& =-\frac{2R}{1∕(\mathit{j\omega C})}=-2\mathit{j\omega RC}=...& \\ & H_3\mathit{j\omega })& =-\frac{R∕2}{\mathit{j\omega L}}=-\frac{1}{2\mathit{j\omega L}∕R}=...& \end{array}$$

These transfer functions should replace their original version in

$$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega L}∕R)\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \end{array}$$

And it can be derived that also after replacing, the circuit will still not oscillate. Any good proof is OK. This can be done by showing the equating the resulting $A_{\mathit{loop}}$ cannot be accomplished for any non-zero finite (radian) frequency.

IF you swap in multiple stages the circuit may get to oscillate!

Q 9.10

(a)

(b)

An answer:
$$\begin{array}{llll}\hfill & H_1(\mathit{j\omega })=A_v\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_2(\mathit{j\omega })=\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_3(\mathit{j\omega })=\frac{1}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & H_4(\mathit{j\omega })=1& \hfill & \\ \hfill & & \hfill \end{array}$$

(c)

An answer:

For these plots, ${\omega }_0=1$. Asymptotic the $|H(j{\omega }_0)|=1=0\mathit{dB}$.

(d)

An answer:
This loopgain has $180^{\circ }$ phase shift at $\omega =0$ and has $-90^{\circ }$ phase shift at $\omega \to \infty$. Somewhere in the middle, close to ${\omega }_0$, the phase shift of the loopgain is $0^{\circ }$. Using a well selected $A_v>1$ the circuit can oscillate harmonically.

(e)

An answer:
$$\begin{array}{llll}\hfill H_{\mathit{loop}}(\mathit{j\omega })& =A_v\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}\cdot \frac{1}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & =A_v\cdot \frac{(\mathit{j\omega L}∕R)\cdot (\mathit{j\omega L}∕R)}{(1+\mathit{j\omega L}∕R)\cdot (1+\mathit{j\omega L}∕R)\cdot (1+\mathit{j\omega RC})}& \hfill & \end{array}$$

This has a real numerator and a complex denominator. If the denominator is real for a specific non-zero and finite $\omega$ then the circuit can oscillate harmonically.

$$\begin{array}{llll}\hfill H_{\mathit{loop}}(\mathit{j\omega })& =A_v\cdot \frac{(\mathit{j\omega L}∕R)\cdot (\mathit{j\omega L}∕R)}{1+\mathit{j\omega }(2L∕R+\mathit{RC})+j^2{\omega }^2(2L^2∕R^2+\mathit{LC})+j^3{\omega }^3{L}^{2}C∕R}& \hfill & \end{array}$$

This is obviously possible for the derived loopgain.

(f)

An answer:
$$\begin{array}{llll}\hfill & \mathit{j\omega }(2L∕R+\mathit{RC})+j^3{\omega }^3{L}^{2}C∕R=0\to & \hfill & \\ \hfill & {\omega }_{\mathit{osc}}=0(\mathrm{\text{no oscillation}})& \hfill & \\ \hfill & {\omega }_{\mathit{osc}}^2=\frac{2L∕R+\mathit{RC}}{L^{2}C∕R}(\mathrm{\text{can oscillate}})& \hfill & \end{array}$$

Substitute this in the next equation to get the required value of $A_v$:

$$\begin{array}{llll}\hfill A_v& =\frac{{\omega }_{\mathit{osc}}^2(2L^2∕R^2+\mathit{LC})-1}{({\omega }_{\mathit{osc}}L∕R)\cdot ({\omega }_{\mathit{osc}}L∕R)}& \hfill & \end{array}$$

(g)

An answer:
Not applicable.

Q 9.11

(a)

(b)

An answer:
$$\begin{array}{llll}\hfill & H_1(\mathit{j\omega })=A_v\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_2(\mathit{j\omega })=\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_3(\mathit{j\omega })=\frac{1}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & H_4(\mathit{j\omega })=\frac{1}{1+\mathit{j\omega RC}}& \hfill & \end{array}$$

(c)

An answer:
$$\begin{array}{llll}\hfill H_{\mathit{loop}}(\mathit{j\omega })& =A_v\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}\cdot \frac{1}{1+\mathit{j\omega RC}}\cdot \frac{1}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & =A_v\cdot \frac{(\mathit{j\omega L}∕R)\cdot (\mathit{j\omega L}∕R)}{(1+\mathit{j\omega L}∕R)\cdot (1+\mathit{j\omega L}∕R)\cdot (1+\mathit{j\omega RC})\cdot (1+\mathit{j\omega RC})}& \hfill & \end{array}$$

This has a real numerator and a complex denominator. If the denominator is real for a specific non-zero and finite $\omega$ then the circuit can oscillate harmonically.

$$\begin{array}{llll}\hfill H_{\mathit{loop}}(\mathit{j\omega })& =\frac{A_v\cdot (\mathit{j\omega L}∕R)\cdot (\mathit{j\omega L}∕R)}{1+2\mathit{j\omega }(\frac{L}{R}+\mathit{RC})+j^2{\omega }^2(\frac{L^2}{R^2}+4\mathit{LC}+R^2{C}^2)+2j^3{\omega }^3(\frac{L^{2}C}{R}+L{C}^{2}R)+j^4{\omega }^4{L}^2{C}^2}& \hfill & \end{array}$$

This is obviously possible for the derived loopgain.

(d)

An answer:
For this, the denominator of the loop gain needs to be set to a real value for a finite, non-zero $\omega$. This follows from solving the equation $$\begin{array}{lll}\hfill 2(\frac{L}{R}+\mathit{RC})-j{\omega }^2\cdot 2(\frac{L^{2}C}{R}+L{C}^{2}R)=0& & \hfill \end{array}$$

This is obviously possible for the derived loopgain.

(e)

An answer:
Not applicable.

Q 9.12

(a)

(b)

An answer:
The combined voltage transfers of the four stages is $A_{\mathit{loop}}(\mathit{j\omega })=\frac{A_{\mathit{buf}}}{{(1+\mathit{j\omega RC})}^4}$

(c)

An answer:

(d)

An answer:
To get $\angle A_{\mathit{loop}}=N\cdot 360^{\circ }$ with a positive $A_{\mathit{buf}}$ two solutions follow:

$\omega =0$
$\omega \to \infty$
These are not considered as valid frequencies for an harmonic oscillator. Hence: no.

(e)

An answer:
To get $\angle A_{\mathit{loop}}=N\cdot 360^{\circ }$ with a negative $A_{\mathit{buf}}$ one solution follows:

$\angle A_{\mathit{loop}}(j{\omega }_{\mathit{osc}})=N\cdot 360^{\circ }$
$\angle \frac{1}{1+\mathit{j\omega RC}}=-45^{\circ }$
this leads to ${\omega }_{\mathit{osc}}=1∕\mathit{RC}$ and $A_{\mathit{buf}}=-4$.

Q 9.13

(a)

(b)

An answer:
The combined voltage transfers of the four stages is $A_{\mathit{loop}}(\mathit{j\omega })=A_{\mathit{buf}}\frac{{(\mathit{j\omega RC})}^4}{{(1+\mathit{j\omega RC})}^4}$

(c)

An answer:

(d)

An answer:
No, the only solutions for $\angle A_{\mathit{loop}}(\mathit{j\omega })$ is for zero or infinite frequencies. Neither of these is considered to be a valid oscillation frequency.

(e)

An answer:
Yes, harmonic oscillation is possible. A derivation using symmetry is the easiest, but plain mathematical solvinf is only a little more work. Both lead to ${\omega }_{\mathit{osc}}=1∕\mathit{RC}$ and $A_{\mathit{buf}}=-4$.

Q 9.14

(a)

(b)

An answer:

$$\begin{array}{llll}\hfill & H_1(\mathit{j\omega })=\frac{1+2\mathit{j\omega L}∕R}{\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_2(\mathit{j\omega })=\frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}& \hfill & \\ \hfill & H_3(\mathit{j\omega })=-\frac{R_1}{R}& \hfill & \end{array}$$

(c)

An answer:

$$\begin{array}{llll}\hfill & -90^{\circ }\le \angle H_1(\mathit{j\omega })\le 0^{\circ }& \hfill & \\ \hfill & -90^{\circ }\le \angle H_2(\mathit{j\omega })\le 0^{\circ }& \hfill & \\ \hfill & \angle H_3(\mathit{j\omega })=\pm 180^{\circ }& \hfill & \end{array}$$

(d)

An answer:

The loop gain is $A_{\mathit{loop}}(\mathit{j\omega })=-\frac{1+2\mathit{j\omega L}∕R}{\mathit{j\omega L}∕R}\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{R_1}{R}$

It can be shown mathematically that $\angle A_{\mathit{loop}}$ can be $N\cdot 360^{\circ }$ only at $\omega =0$. This is not considered as a valid oscillation frequency.

(e)

An answer:

$$\begin{array}{llll}\hfill & H_1(\mathit{j\omega })=\frac{1+2\mathit{j\omega L}∕R}{\mathit{j\omega L}∕R}& \hfill & \\ \hfill & H_2(\mathit{j\omega })=\frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}& \hfill & \\ \hfill & H_3(\mathit{j\omega })=-\frac{1}{\mathit{j\omega RC}}& \hfill & \end{array}$$

(f)

An answer:

$$\begin{array}{llll}\hfill & -90^{\circ }\le \angle H_1(\mathit{j\omega })\le 0^{\circ }& \hfill & \\ \hfill & -90^{\circ }\le \angle H_2(\mathit{j\omega })\le 0^{\circ }& \hfill & \\ \hfill & \angle H_3(\mathit{j\omega })=90^{\circ }& \hfill & \end{array}$$

(g)

An answer:

The loop gain is $A_{\mathit{loop}}(\mathit{j\omega })=-A_{\mathit{buf}}\cdot \frac{1+2\mathit{j\omega L}∕R}{\mathit{j\omega L}∕R}\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{1}{\mathit{j\omega RC}}$

The denominator is imaginary. If the numerator can be set to an imaginary value for a non-zero finite $\omega$ then this circuit can be dimensioned to oscillate harmonically. Working things out a little leads to:

$$\begin{array}{llll}\hfill & A_{\mathit{loop}}(\mathit{j\omega })=-A_{\mathit{buf}}\cdot \frac{1+\mathit{j\omega }(2L∕R+\mathit{RC})+2{(\mathit{j\omega })}^2\mathit{LC}}{{(\mathit{j\omega })}^{3}L{C}^{2}R}& \hfill & \\ \hfill & {\omega }_{\mathit{osc}}=\frac{1}{\sqrt{2\mathit{LC}}}& \hfill & \\ \hfill & A_{\mathit{buf}}=\frac{\mathit{RC}}{2\mathit{RC}+4L∕R}& \hfill & \end{array}$$

Q 9.15

(a)

The circuit can be decomposed into 3 stages that have frequency dependent transfer functions (identified by their gray background in the figure), and one ideal voltage buffer stage with a real valued voltage gain $A_{\mathit{buf}}$.

(b)

An answer:
see 6, question 7.

(c)

An answer:

$$\begin{array}{lllllll}\hfill & \angle H_1(j0)=0^{\circ }& \hfill \angle H_1(j\infty )=-180^{\circ }& & \hfill & & \hfill \\ \hfill & \angle H_2(j0)=0^{\circ }& \hfill \angle H_2(j\infty )=-180^{\circ }& & \hfill & & \hfill \\ \hfill & \angle H_3(j0)=90^{\circ }& \hfill \angle H_3(j\infty )=0^{\circ }& & \hfill & & \hfill \end{array}$$

(d)

An answer:
Using the specified R and C and using $\omega =1∕\mathit{RC}$:

(e)

An answer:

The loop gain is $A_{\mathit{loop}}(\mathit{j\omega })=A_{\mathit{buf}}\cdot \frac{1-\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{1-\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}$. Working on this equation can be done of course. But, the explicit hint — both of them — simplifies a lot.

Using the phases of the three stages, it follows that

• $\angle A_{\mathit{loop}}(0)=90^{\circ }$
• $\angle A_{\mathit{loop}}(\infty )=-360^{\circ }$
• the total phase shift ranges from $90^{\circ }\to -360^{\circ }$
• the phase shift (change) of the first two stages it twice that of the third stage.

It follows that for $A_{\mathit{buf}}>0$ there are two possibilities to get $\angle A_{\mathit{loop}}=N\cdot 360^{\circ }$:

• $\omega \to \infty$
• some finite $\omega$ close to the ${\omega }_0$ in the Bode plot.

There is no way to get harmonic oscillation at the finite $\omega$ solution because the $|A_{\mathit{loop}}|$ is higher at $\omega \to \infty$ and hence the solution at that $\omega \to \infty$ will win. And that’s not regarded as oscillation.

IF you would calculate the finite $\omega$ solution, then the answer would follow from having to achieve $-90^{\circ }$ phase shift in the 3 stages, resulting in respectively (symmetry!) $-90^{\circ }\ast 2∕5=-36^{\circ }$, $-90^{\circ }\ast 2∕5=-36^{\circ }$, $-90^{\circ }\ast 1∕5=-18^{\circ }$ phase shift. That would then result in ${\omega }_{\mathit{osc}}=1∕\mathit{RC}\cdot \mathit{tan}(18^{\circ })$. But, sorry, the $\omega \to \infty$ wins...

Using the other hint requires rewriting transfer functions to standard form. The three forms would then be:

$$\begin{array}{llll}\hfill & H_{1,2}(\mathit{j\omega })=\frac{1-\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\to \frac{1+{(\mathit{\omega RC})}^2}{{(1+\mathit{j\omega RC})}^2}& \hfill & \\ \hfill & H_3(\mathit{j\omega })=\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}& \hfill & \end{array}$$

And the loop gain would then be: $A_{\mathit{loop}}(\mathit{j\omega })=A_{\mathit{buf}}\cdot \frac{1+{(\mathit{\omega RC})}^2}{{(1+\mathit{j\omega RC})}^2}\cdot \frac{1+{(\mathit{\omega RC})}^2}{{(1+\mathit{j\omega RC})}^2}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}$ which has an imaginary numerator and a complex denominator. This enables relatively easy solving, leading to the exact same findings as before.

(f)

An answer:

The same analyses as for the previous question can be used.

Using the phases of the three stages and including the $\pm 180^{\circ }$ of $A_{\mathit{buf}}$, it follows that

• $\angle A_{\mathit{loop}}(0)=270^{\circ }$
• $\angle A_{\mathit{loop}}(\infty )=-180^{\circ }$
• the total phase shift ranges from $270^{\circ }\to -180^{\circ }$
• the phase shift (change) of the first two stages it twice that of the third stage.

It follows that for $A_{\mathit{buf}}<0$ there is only one possibility to get $\angle A_{\mathit{loop}}=N\cdot 360^{\circ }$:

• some finite $\omega$ close to the ${\omega }_0$ in the Bode plot.

For this solution, the 3 stages combined have to achieve $-270^{\circ }$ phase shift, resulting in respectively (symmetry!) $-270^{\circ }\ast 2∕5=-108^{\circ }$, $-270^{\circ }\ast 2∕5=-108^{\circ }$, $-270^{\circ }\ast 1∕5=-54^{\circ }$ phase shift. That would then result in ${\omega }_{\mathit{osc}}=1∕\mathit{RC}\cdot \mathit{tan}(54^{\circ })$. $A_{\mathit{buf}}$ then follows from equating $A_{\mathit{loop}}(j{\omega }_{\mathit{osc}})=1$

Q 9.16

(a)

(b)

(c)

(d)

Q 9.17

(a)

An answer:
The closed-loop transfer is $H(\mathit{j\omega })=\frac{A}{1+A\cdot \beta (\mathit{j\omega })}$. For $A\cdot \beta (\mathit{j\omega })=1$, $\frac{A}{1+A\cdot \beta (\mathit{j\omega })}\to \infty$, hence a sine would be sustained indefinitely without any input signal. This is harmonic oscillation, with in equation $v_{\mathit{out}}=H(\mathit{j\omega })\cdot v_{\mathit{in}}=\infty \cdot 0$.

(b)

An answer:
The derivation of $H(\mathit{j\omega })$ can be done in many ways. One of these is shown below. $$\begin{array}{rcll}& v_{\mathit{out}}& =A(v_{+}-v_{-})& \\ & & v_{-}=v_{\mathit{out}}& \\ & & v_{+}=v_x\frac{1}{1+\mathit{j\omega RC}}& \\ & & v_{-}=v_{+}(A\to \infty )& \end{array}$$

after which $v_x$ can be calculated in a variety of ways. E.g. applying KCL on node $v_x$ gives:

$$\frac{v_{\mathit{in}}-v_x}{R}=\frac{v_x-v_{\mathit{out}}}{1∕\mathit{j\omega C}}+\frac{v_x-v_{+}}{R}$$

Rewriting $v_{+}=v_x\frac{1}{1+\mathit{j\omega RC}}$ into $v_x=(1+\mathit{j\omega RC})v_{\mathit{out}}$ and using that along with $v_{+}=v_{\mathit{out}}$ yields

$$\begin{array}{rcll}& & v_{\mathit{out}}=\frac{v_{\mathit{in}}+\mathit{j\omega RC}\cdot v_{\mathit{out}}}{1+3\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2}& \\ & & (1+3\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2)v_{\mathit{out}}=v_{\mathit{in}}+\mathit{j\omega RC}\cdot v_{\mathit{out}}& \\ & H(\mathit{j\omega })& =\frac{1}{1+2\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2}& \end{array}$$

(c)

An answer:
One of the many correct derivation of $H(\mathit{j\omega })$ is shown below.

$$\begin{array}{rcll}& v_{\mathit{out}}& =A(v_{+}-v_{-})& \\ & & v_{-}=v_{\mathit{in}}\cdot \frac{\mathit{j\omega L}}{1∕\mathit{j\omega C}+\mathit{j\omega L}}+v_{\mathit{out}}\cdot \frac{1∕\mathit{j\omega C}}{1∕\mathit{j\omega C}+\mathit{j\omega L}}& \\ & & v_{+}=v_{\mathit{in}}\frac{\mathit{j\omega L}}{R+\mathit{j\omega L}}& \\ & & v_{-}=v_{+}(A\to \infty )& \end{array}$$

Yielding

$$\begin{array}{rcll}& & v_{\mathit{in}}\cdot \frac{j^2{\omega }^2\mathit{LC}}{1+j^2{\omega }^2\mathit{LC}}+v_{\mathit{out}}\cdot \frac{1}{1+j^2{\omega }^2\mathit{LC}}=v_{\mathit{in}}\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \\ & & v_{\mathit{out}}\cdot \frac{1}{1+j^2{\omega }^2\mathit{LC}}=v_{\mathit{in}}\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}-v_{\mathit{in}}\cdot \frac{j^2{\omega }^2\mathit{LC}}{1+j^2{\omega }^2\mathit{LC}}& \\ & H(\mathit{j\omega })& =\frac{\mathit{j\omega L}∕R-j^2{\omega }^2\mathit{LC}}{1+\mathit{j\omega L}∕R}& \end{array}$$

(d)

An answer:

$$\begin{array}{rcll}& & H_1=\frac{1}{{(1+\mathit{j\omega RC})}^2}(\mathit{see}\mathit{before})& \\ & & H_2=\frac{\mathit{j\omega }\frac{L}{R}(1-\mathit{j\omega RC})}{\mathit{j\omega }\frac{L}{R}+1}(\mathit{see}\mathit{before})& \\ & & H_3=-\frac{R}{R+\mathit{j\omega L}}=-\frac{1}{1+\mathit{j\omega }\frac{L}{R}}& \end{array}$$

(e)

An answer:

(f)

An answer:
The open-loop transfer function is:

$$A_{\mathit{loop}}(\omega )=A_{\mathit{buf}}\frac{\mathit{j\omega }\frac{L}{R}(1-\mathit{j\omega RC})}{{(1+\mathit{j\omega RC})}^2{(1+\mathit{j\omega }\frac{L}{R})}^2}$$

Assuming the same corner frequencies for the sections,

$$A_{\mathit{loop}}(\omega )=A_{\mathit{buf}}\frac{j\frac{\omega }{{\omega }_c}(1-j\frac{\omega }{{\omega }_c})}{{(1+\mathit{j\omega }\frac{\omega }{{\omega }_c})}^4}$$

(g)

An answer:
yes. $$\begin{array}{rcll}& & {\omega }_{\mathit{osc}}=\mathit{tan}\left(\frac{\pi }{10}\right){\omega }_c& \\ & & A_{\mathit{buf}}=\frac{1+\mathit{tan}\left(\frac{\pi }{10}\right)}{\mathit{tan}\left(\frac{\pi }{10}\right)}& \end{array}$$

(h)

An answer:
yes. $$\begin{array}{rcll}& & {\omega }_{\mathit{osc}}=\mathit{tan}\left(\frac{3\pi }{10}\right){\omega }_c& \\ & & A_{\mathit{buf}}=-\frac{1+\mathit{tan}\left(\frac{3\pi }{10}\right)}{\mathit{tan}\left(\frac{3\pi }{10}\right)}& \end{array}$$

Q 9.18

(a)

(b)

An answer:
(c)

(c)

An answer:
no, $\omega =0$ may result but that is not regarded an harmonic oscillation.

(d)

An answer:
yes, at $\omega =\frac{\mathit{tan}(60^{\circ }}{\mathit{RC}}$