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Course by: Laurence Riddle. E-mail the author

Signal Power

Module by: Don Johnson. E-mail the author

Summary: This module examines signal power, looking at instantaneous and average power. It uses orthogonality properties to derive a simple expression for average power. It also defines and displays a power spectrum.

An interesting question you could ask about a signal is its average power. A signal's instantaneous power is defined to be its square, as if it were a voltage or current passing through a 1 Ω resistor. The average power is the average of the instantaneous power over some time interval. For a periodic signal, the natural time interval is clearly its period; for nonperiodic signals, a better choice would be entire time or time from onset. For a periodic signal, the average power is the square of the root-mean-squared (rms) value. We define the rms value of a periodic signal to be

rmss=21T0Ts2tdt rms s 2 1 T t 0 T s t 2
(1)
and thus its average power is rms2s rms s 2 .
powers=rms2s=1T0Ts2tdt power s rms s 2 1 T t 0 T s t 2
(2)

Exercise 1

What is the rms value of the half-wave rectified sinusoid?

Solution

A half-wave rectified sinusoid has half the average power of the original sine wave since it is zero half the time. A sine wave's average power equals A22 A 2 2 , making the rms value of the half-wave rectified signal A2 A 2 .

To find the average power in the frequency domain, we need to substitute the spectral representation of the signal into this expression.

powers=1T0T a 0 +k=1 a k cos2πktT+k=1 b k sin2πktT2dt power s 1 T t 0 T a 0 k 1 a k 2 k t T k 1 b k 2 k t T 2
(3)
The square inside the integral will contain all possible pairwise products. However, the orthogonality properties say that most of these crossterms integrate to zero. The survivors leave a rather simple expression for the power we seek.
powers= a 0 2+12k=1 a k 2+ b k 2 power s a 0 2 1 2 k 1 a k 2 b k 2
(4)

It could well be that computing this sum is easier than integrating the signal's square. Furthermore, the contribution of each term in the Fourier series toward representing the signal can be measured by its contribution to the signal's average power. Thus, the power contained in a signal at its k k th harmonic is a k 2+ b k 22 a k 2 b k 2 2 . The power spectrum P s k P s k , such as shown in Figure 1, plots each harmonic's contribution to the total power.

Exercise 2

In stereophonic systems, deviation of a sine wave from the ideal is measured by the total harmonic distortion, which equals the total power in the harmonics higher than the first compared to power in the fundamental. Find an expression for the total harmonic distortion for any periodic signal. Is this calculation most easily performed in the time or frequency domain?

Solution

Total harmonic distortion equals k=2 a k 2+ b k 2 a 1 2+ b 1 2 k 2 a k 2 b k 2 a 1 2 b 1 2 . Clearly, this quantity is most easily computed in the frequency domain. However, the numerator equals the the square of the signal's rms value minus the power in the average and the power in the first harmonic.

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