Frequency response from a circuit diagram

In this module we calculate the frequency response from a circuit diagram of a simple analog filter, as shown in . We know that the frequency response, denoted by $(H(i(\Omega )))$ , is calculated as ratio of the output and input voltages (in the frequency domain). That is,

Now, to calculate the frequency response we find expressions for $V_{\mathrm{in}}$ , and $V_{\mathrm{out}}$ , as follows

Implicit in using the transfer function is that the input is a complex exponential, and the output is also a complexexponential having the same frequency. The transfer function reveals how the circuit modifies the input amplitude in creating the output amplitude. Thus, the transfer function completely describes how the circuit processes the input complex exponential to produce the output complex exponential. The circuit's function is thus summarizedby the transfer function. In fact, circuits are often designed to meet transfer function specifications. Because transfer functions are complex-valued, frequency-dependent quantities, we can better appreciate a circuit's function by examining themagnitude and phase of its transfer function ( ). Note that in we plot the magnitude phase as a function of the frequency $F$ , instead of the angular frequency $\Omega$ . Since $\Omega =2\pi F$ , this is just a matter of taste, see Frequency definitions and peridocity for details.

Several things to note about this transfer function.

We can compute the frequency response for both positive and negative frequencies. Recall that sinusoids consist of the sumof two complex exponentials, one having the negative frequency of the other. We will consider how the circuit acts on asinusoid soon. Do note that the magnitude has even symmetry : The negative frequency portion is a mirror image of the positive frequency portion: $\left|H(-(iF))\right|=\left|H(iF)\right|$ . The phase has odd symmetry : $\mathop{\arg }(H(-(iF)))=-\mathop{\arg }(H(iF))$ . These properties of this specific example apply for all transfer functions associated with circuits. Consequently, we don't need to plot the negativefrequency component; we know what it is from the positive frequency part.

The magnitude equals $\frac{1}{\sqrt{2}}$ of its maximum gain (1 at $F=0$ ) when $2\pi FRC=1$ (the two terms in the denominator of the magnitude are equal). The frequency $F_c=\frac{1}{2\pi RC}$ defines the boundary between two operating ranges.

• For frequencies below this frequency, the circuit does not much alter the amplitude of the complex exponentialsource.
• For frequencies greater than $F_c$ , the circuit strongly attenuates the amplitude. Thus, when the source frequency is in this range, thecircuit's output has a much smaller amplitude than that of the source.

The phase shift caused by the circuit at the cutoff frequencyprecisely equals $-\left(\frac{\pi }{4}\right)$ . Thus, below the cutoff frequency, phase is little affected, but athigher frequencies, the phase shift caused by the circuit becomes $-\left(\frac{\pi }{2}\right)$ . This phase shift corresponds to the difference between a cosine and a sine.

We can use the transfer function to find the output when theinput voltage is a sinusoid for two reasons. First of all, a sinusoid is the sum of two complex exponentials, each having afrequency equal to the negative of the other. Secondly, because the circuit is linear, superposition applies. If the source isa sine wave, we know that

The notion of impedance arises when we assume the sources arecomplex exponentials. This assumption may seem restrictive; what would we do if the source were a unit step? When we useimpedances to find the transfer function between the source and the output variable, we can derive from it the differentialequation that relates input and output. The differential equation applies no matter what the source may be. As we have argued, it isfar simpler to use impedances to find the differential equation (because we can use series and parallel combination rules) thanany other method. In this sense, we have not lost anything by temporarily pretending the source is a complex exponential.

In fact we can also solve the differential equation usingimpedances! Thus, despite the apparent restrictiveness of impedances, assuming complex exponential sources is actuallyquite general.