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Transfer Function of a System with Sinusoidal Input

Module by: Don Johnson. E-mail the author

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Summary: Discussing the special case of Transfer Function, that is when the input voltage is a sinusoid.

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One very important special case of transfer functions occurs when the input voltage is a sinusoid. Because a sinusoid is the sum of two complex exponentials, each having a frequency equal to the negative of the other, we can directly predict the output voltage by examining the transfer function. If the source is a sine wave, we know that

v in t=Asin2πft=A2ⅈⅇⅈ2πft−ⅇ-ⅈ2πft v in t A 2 f t A 2 2 f t 2 f t (1)
Since the input is the sum of two complex exponentials, we know that the output is also a sum of two similar complex exponentials, the only difference being that the complex amplitude of each is multiplied by the transfer function evaluated at each exponential's frequency.
v out t=A2ⅈHfⅇⅈ2πft−A2ⅈH-fⅇ-ⅈ2πft v out t A 2 H f 2 f t A 2 H f 2 f t (2)
As noted earlier, the transfer function is most conveniently expressed in polar form: Hf=|Hf|ⅇⅈ∠Hf H f H f H f . Furthermore, |H-f|=|Hf| H f H f (even symmetry of the magnitude) and ∠H-f=-∠Hf H f H f (odd symmetry of the phase). The output voltage expression simplifies to
v out t=A2ⅈ|Hf|ⅇⅈ2πft+∠Hf−A2ⅈ|Hf|ⅇ-ⅈ2πft−∠Hf=A|Hf|sin2πft+∠Hf v out t A 2 H f 2 f t H f A 2 H f 2 f t H f A H f 2 f t H f (3)
The circuit's output to a sinusoidal input is also a sinusoid, having a gain equal to the magnitude of the circuit's transfer function evaluated at the source frequency and a phase equal to the phase of the transfer function at the source frequency. It will turn out that this input-output relation description applies to any linear circuit having a sinusoidal source.

Exercise 1

This input-output property is a special case of a more general result. Show that if the source can be written as the imaginary part of a complex exponential— v in t=ℑVⅇⅈ2πft v in t V 2 f t —the output is given by v out t=ℑVHfⅇⅈ2πft v out t V H f 2 f t . Show that a similar result also holds for the real part.

Solution

The key notion is writing the imaginary part as the difference between a complex exponential and its complex conjugate:

ℑVⅇⅈ2πft=Vⅇⅈ2πft−V¯ⅇ-ⅈ2πft2ⅈ V 2 f t V 2 f t V 2 f t 2 (4)
The response to Vⅇⅈ2πft V 2 f t is VHfⅇⅈ2πft V H f 2 f t , which means the response to V¯ⅇ-ⅈ2πft V 2 f t is V¯H-fⅇ-ⅈ2πft V H f 2 f t . as H-f=H¯f H f H f , the Superposition Principle says that the output to the imaginary part is ℑVHfⅇⅈ2πft V H f 2 f t . The same argument holds for the real part: ℜVⅇⅈ2πft→ℜVHfⅇⅈ2πft V 2 f t V H f 2 f t .

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