Given. Ideal inverting op-amp; input impedance $Z_1=R_1+1/(j\omega C_1)$ ($R_1=10\text{k}\Omega$, $C_1=0.1\mu\text{F}$); feedback $Z_f=R_2=100\text{k}\Omega$. $V_i(t)=10\sin(120\pi t)$ V for parts (c),(d).
Given data
Quantity
Value
$R_1$
$10\text{k}\Omega$
$C_1$
$0.1\mu\text{F}$
$R_2$
$100\text{k}\Omega$
$V_i(t)$ (c,d)
$10\sin(120\pi t)$ V, $f=60\text{Hz}$
Find. $H(j\omega)=V_o/V_i$; $f_{3dB}$ and its Bode sketch; $V_o(j\omega)$ and $V_o(t)$ at 60Hz.
Approach. Use the ideal-op-amp inverting-amplifier relation $V_o/V_i=-Z_f/Z_1$ with the series $R_1$-$C_1$ input impedance; the resulting single-pole high-pass shape gives $f_{3dB}$ directly from $R_1C_1$, and the passband gain from $R_2/R_1$.
Part (a) — transfer function. Virtual ground at the inverting input, ideal op-amp (infinite gain, no input current):
$$\frac{V_o}{V_i}=-\frac{Z_f}{Z_1}=-\frac{R_2}{R_1+\dfrac{1}{j\omega C_1}}=\boxed{\dfrac{-j\omega R_2C_1}{1+j\omega R_1C_1}}$$
This is a single-pole high-pass response: at low $\omega$, $H\to0$ (capacitor blocks); at high $\omega$, $H\to-R_2/R_1$ (capacitor shorts, plain inverting amp).
Part (b) — 3dB frequency and sketch. The pole occurs where $\omega R_1C_1=1$:
$$f_{3dB}=\frac{1}{2\pi R_1C_1}=\frac{1}{2\pi(10^4)(10^{-7})}=\boxed{159\text{ Hz}}$$
Passband gain (well above $f_{3dB}$): $|H|_{\max}=R_2/R_1=\boxed{10\text{ V/V} = 20\text{ dB}}$. Below $f_{3dB}$ the magnitude falls at $+20\text{dB/decade}$ (numerator $\propto\omega$ dominates); above it, flattens to the 20dB plateau.
Part (c) — $V_o(j\omega)$ at 60Hz. $\omega=2\pi(60)=120\pi=377\text{ rad/s}$. Then $\omega R_1C_1=0.377$, $\omega R_2C_1=3.77$:
$$H(j\omega)=\frac{-j(3.77)}{1+j(0.377)}=3.53\angle{-110.7^\circ}$$
With $V_i$ phasor amplitude $10\text{V}\angle0^\circ$:
$$V_o(j\omega)=10\times3.53\angle{-110.7^\circ}=\boxed{35.3\angle{-110.7^\circ}\text{ V}}$$
Part (d) — $V_o(t)$. Converting the phasor back to the time domain at $\omega=120\pi$:
$$V_o(t)=\boxed{35.3\sin(120\pi t-110.7^\circ)\text{ V}}$$
Fig. Q3(b) — Bode magnitude sketch. Rises at +20dB/decade below f3dB=159Hz (high-pass corner set by R1C1), flattens to the passband gain 20log₁₀(R2/R1)=20dB above it.
Final Results — Question 3
Quantity
Value
$H(j\omega)$
$-j\omega R_2C_1/(1+j\omega R_1C_1)$
$f_{3dB}$
$159\text{ Hz}$
Passband gain
$10\text{ V/V}$ (20dB)
$V_o(j\omega)$ at 60Hz
$35.3\angle{-110.7^\circ}\text{ V}$
$V_o(t)$ at 60Hz
$35.3\sin(120\pi t-110.7^\circ)\text{ V}$
Check — ideal-op-amp linear model. The computed $35.3\text{V}$ output peak assumes an ideal op-amp with unlimited supply rails; a real device would clip at its supply rails long before reaching this amplitude. The question specifies an ideal op-amp, so the linear-model answer is reported as derived.
