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Impedance Example

Module by: Don Johnson

Summary: Given an impedance Example to see how to use impedance.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

To illustrate the impedance approach, we refer to the R C RC circuit (Figure 1) below, and we assume that v in = V in ⅇⅈ2πft v in V in 2 f t .

Figure 1: A simple R C RC circuit.
(circuit5.png)

Using impedances, the complex amplitude of the output voltage V out V out can be found using voltage divider:

V out = Z C Z C + Z R V in V out Z C Z C Z R V in (1)
V out =1ⅈ2πfC1ⅈ2πfC+R V in V out 1 2 f C 1 2 f C R V in (2)
V out =1ⅈ2πfRC+1 V in V out 1 2 f R C 1 V in (3)
If we refer to the differential equation for this circuit (shown in Circuits with Capacitors and Inductors to be RCddt V out + V out = V in R C t V out V out V in ), letting the output and input voltages be complex exponentials, we obtain the same relationship between their complex amplitudes. Thus, using impedances is equivalent to using the differential equation and solving it when the source is a complex exponential.

In fact, we can find the differential equation directly using impedances. If we cross-multiply the relation between input and output amplitudes,

V out ⅈ2πfRC+1= V in V out 2 f R C 1 V in (4)
and then put the complex exponentials back in, we have
RCⅈ2πf V out ⅇⅈ2πft+ V out ⅇⅈ2πft= V in ⅇⅈ2πft R C 2 f V out 2 f t V out 2 f t V in 2 f t (5)
In the process of defining impedances, note that the factor ⅈ2πf 2 f arises from the derivative of a complex exponential. We can reverse the impedance process, and revert back to the differential equation.
RCddt V out + V out = V in R C t V out V out V in (6)
This is the same equation that was derived much more tediously in Circuits with Capacitors and Inductors. Finding the differential equation relating output to input is far simpler when we use impedances than with any other technique.

Exercise 1

Suppose you had an expression where a complex amplitude was divided by ⅈ2πf 2 f . How did this division arise?

Solution

Division by ⅈ2πf 2 f arises from integrating a complex exponential. Consequently,

1ⅈ2πfV⇔∫Vⅇⅈ2πftdt ⇔ 1 2 f V t V 2 f t (7)

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