Fourier Series Application

For sinusoidal inputs we can already use phasor analysis. But circuits may often have inputs that are nonsinusoidal periodic functions. We can use the Fourier Series to break down these periodic functions into sinusoids and find the steady state response by using phasor analysis and superposition.

Steps for Applying Fourier Series

1. Express the input as a Fourier series.

2. Transform the circuit from the time domain to the frequency domain (phasor domain, not s domain).

3. Find the response of the DC and AC components in the Fourier series.

4. Add the individual DC and AC responses using the superposition principle.

Essentially what we are doing is using the Fourier series to express the periodic signal input as a sum of sinusoids.

Superposition(which you learned in your previous Circuits class) says we can analyze each input separately, calculate each output separately, then add all the outputs together to get the final output due to the entire input.

Example:

Suppose we have a circuit and input f(t) shown below. Find the response Vo(t). So find Vo(t) when Vs(t) = f(t) shown below.

Notice the input is not sinusoidal so we cannot use phasor analysis here. We have already solve it's Fourier Series previously as:

Step 1 is already done for us. We have solved the Fourier series.

Step 2 is convert the circuit into the phasor domain

• In the phasor domain we can solve the general output using voltage division

Step 3 is to find the outputs of all the components in the Fourier Series Separately.

The DC Component

For the DC component, 𝑛 = 0 so Vo = 0. (We could also get there by noting that when a DC source is applied, an inductor looks like a short circuit and a capacitor looks like an open circuit.)

The AC Components

Extra analysis

• We see that the above circuit is a high pass filter. (circuit analysis or analyzing the inductor)

• The output is shown above mathematically but let's write the first 3 terms to see what is happening. Let's write the first 3 terms of the input as well.

Vs(t) = 0.6369 cos ( 𝜋t -90° ) + 0.2123 cos ( 3𝜋t -90° ) + 0.1273 cos ( 5𝜋t -90° )

Vo(t) = 0.4981 cos ( 𝜋t -51.49° ) + 0.2051 cos ( 3𝜋t -75.14° ) + 0.1257 cos ( 5𝜋t -80.96° )

You can see above that the lowest frequency term is attenuated by a larger factor while the highest frequency term is attenuated by a smaller factor. The lowpass filter is letting the signals with higher frequency flow to the output without attenuation and is blocking the lower frequency signals by attenuating their magnitudes. You should also notice the phase shifts change as frequencies change. This is all consistent with what we've learned about phasor analysis and Transfer Functions.