Solving Two Ways

To start, let's consider an example circuit we've seen a few times:

Here, we'll assume that the input V_{\rm I}(t) is sinusoidal, i.e., that V_{\rm I}(t) = A\cos(\omega t + \phi_a). We'll also assume that the output takes the form of v_{\rm C}(t) = B\cos(\omega t + \phi_b). We'll solve this circuit two ways here, and they should give us the same result.

1) Differential Equations

1. Using methods we've seen before, we can write a differential equation that relates V_{\rm I}(t) to v_{\rm C}(t). Go ahead and do that.

2. We'll temporarily create a new circuit where we assume that the input is instead given by \tilde{V}_{\rm I}(t) = \tilde{A}e^{j\omega t}, where \tilde{A} is a complex number such that the real part of \tilde{V}_{\rm I}(t) is our original cosine function, i.e.:

   {\rm Re}\left(\tilde{V}_{\rm I}(t)\right) = {\rm Re}\left(\tilde{A}e^{j\omega t}\right) = A\cos(\omega t + \phi_a)

   In order for this to be the case, what must \tilde{A} be? Express your answer in polar form.

3. Assuming this complex-valued input V_{\rm I}(t), assume that the voltage across the capacitor will also be some value \tilde{v}_{\rm C}(t) = \tilde{B}e^{j\omega t}, where \tilde{B} may be complex-valued (and note that the answer to our original problem, v_{\rm C}(t), will be the real part of \tilde{v}_{\rm C}(t)).

   Use the differential equation above to find \tilde{B}. What are its magnitude and phase as functions of the circuit parameters \tilde{A} and \omega?

4. Using that result, let's return to our original question and find v_{\rm C}(t) = B\cos(\omega t + \phi_b). Express B and \phi_b in terms of the circuit parameters A, \omega, and \phi_a.

5. We will often characterize a system by the ratio of \tilde{B} and \tilde{A}, viewed as a function of the frequency \omega. We call this representation the frequency response of the circuit, often denoted as \tilde{H}(\omega). If the input to a circuit is a sinusoid at a certain frequency, the output will be a sinusoid at the same frequency; the frequency response tells us how the amplitude and phase of that output sinusoid relate to those of the input.

   Sketch plots of the magnitude and angle of \tilde{H}(\omega) as functions of \omega. Recall that we characterized this circuit as a low-pass filter in lab, asserting that it attenuated high frequencies heavily, while leaving low frequencies roughly unchanged. Explain how this manifests in your sketches.

2) Impedance

Now let's try to solve this circuit using impedance instead. Our first step will be to replace all of the components with equivalent impedances:

1. Given the components in the original circuit, what are the impedances \tilde{Z}_1 and \tilde{Z}_2?

2. In this arrangement (and, again, under the assumption that \tilde{V}_{\rm I}(t) = \tilde{A}e^{j\omega t}, this now looks like a voltage divider, so we can solve using our trusty voltage divider equation: \tilde{v}_{\rm C}(t) = \frac{\tilde{Z}_2}{\tilde{Z}_1 + \tilde{Z}_2}\tilde{V}_{\rm I}(t).

3. Solve things out this way and relate your answer back to the one you found in the last part (using differential equations directly).

 
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