# Noise. Noise is like a weed. Just as a weed is a plant where you do not wish it to be, noise is a signal where you do not wish it to be. The noise signal.

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Noise

Noise is like a weed. Just as a weed is a plant where you do not wish it to be, noise is a signal where you do not wish it to be. The noise signal interferes with the signals of interest. If we knew exactly what the interfering noise signal was, we could simply subtract it from the original signal. (Assuming that the noise is additive.) Since we generally do not know what the noise signal is, we must know something about its nature in order to attenuate its interfering effect on other signals.

Power Spectral Density One thing that we typically know about noise is something about its spectrum. If the noise signal is n(t), the noise spectrum is N(f).

Instead of working with the noise spectrum [N(f)], we typically work with the noise power spectral density. The notation for the power spectral density of the noise n(t) is

The power spectral density of the noise is the power per Hz of the noise. Dimensionally, Power = (Power Spectral Density) x (Bandwidth). In general, when the power spectral density is not constant, we can find the power from the power spectral density using

Example: Find the power when the power spectral density is equal to Solution:

Example: Suppose the noise in the previous example is passed through a low-pass filter with cutoff frequency of 10 Hz. What is the output noise power? Solution:

Example: The noise power spectral density is This noise is passed through a filter whose bandwidth is B. Find the output noise power.

Solution: This kind of noise has a (power) spectrum which is constant over all frequencies. For this reason, the noise is called white noise.

Example: Find the power when the power spectral density is equal to Solution:

This integral can be evaluated using trigonometric substitution. Let f = tan .

Example: The noise power spectral density is This noise is passed through a filter whose bandwidth is B and whose voltage gain is G. Find the output noise power.

Solution: We proceed as before, but we introduce a factor proportional to the power gain of the signal. Power Gain = [Voltage Gain] 2 For this example, the power gain is G 2. The power spectral density at the output of the filter is N 0 G 2 /2. Thus the power is

Exercise: White noise whose power spectral density is N 0 /2 is input to a low-pass filter whose transfer function (gain versus frequency) is Find the output noise power. The gain-squared will be H(f)H*(f). Use the results of previous integrations in this lecture.

Given the spectrum of a signal, the power may be computed using one of two forms of a theorem called Parseval’s theorem. In future calculations, the spectrum of a signal or the spectrum of noise will be given, an the powers will need to be computed from these spectra.

Parseval’s Theorem for Fourier Series Given a signal described by a Fourier series, the power can be found from its Fourier series coefficients by using something called Parseval’s Theorem for Fourier series.

Let x(t) be a periodic signal described by a Fourier series: The instantaneous power is proportional to x 2 (t):

This instantaneous power can also be expressed as The term X m * is the complex conjugate of X m. (The magnitude-squared of any complex number is given by |x| 2 =xx*.)

This instantaneous power can now be expressed as To find the average power, we integrate x 2 (t) and divide by the time interval.

Inserting the expression for x 2 (t) into the integral for average power, we have

It can be shown that (exercise for the student): This function acts like a selector function sifting out those terms for which n=m. So,

Example: Using Parseval’s theorem for Fourier series, find the power in a simple sinewave Solution: First, we convert cos  c t to complex exponentials:

Thus we have, X 1 = X -1 =½. The power according to Parseval’s theorem is This result should come as no surprise since the average power is equal to the square of the rms voltage, and the rms voltage is equal to 1/  2.

Example: Find the average power in a squarewave using Parseval’s theorem. Compare the result to finding the average power by finding the average of the square of the squarewave. Solution: The squarewave has the following Fourier series expansion:

Taking the sums of the squares of the Fourier series components, we have

Letting We can calculate P N for various values of N.

The series seems to converge to one. Now we compute the average of the square of the squarewave. In the graph below, the unsquared squarewave is denoted in green; the squared squarewave is denoted in blue (dashed). [(-1) 2 = 1.] 1 x(t)x 2 (t)

We see that the value of the square of the squarewave is one, and that the average value is one.

Exercise: Find the average power in the following signal: Use Parseval’s theorem for Fourier series.

Parseval’s Theorem for Fourier Transforms Parseval’s theorem for Fourier transforms uses Fourier transforms to calculate the energy in a signal. Energy is equal to power times time. If the power is not constant, energy is equal to the integral of the power over time.

Energy = (Power) x (Time).

We can manipulate the integral expression by using Fourier transforms:

So, generally speaking, the integral of the square of the time-domain signal is equal to the integral of the square of the corresponding frequency-domain signal.

Example: Find the energy in the signal first, using the integral of x 2 (t), then using the integral of X 2 (t).

Let 2  f = tan .

We see that Parseval’s theorem gives consistent results in the computation of energy for both the time and the frequency domain.

Exercise: Find the energy in the function Use Parseval’s theorem for Fourier transforms: it is impossible to integrate sinc t analytically.

The Quadrature Decomposition of Noise Very often noise is added to a modulated signal, that is a signal whose frequency components are centered about some high frequency f c. The corresponding noise is usually passed through a band-pass filter in the demodulation process. The noise of interest is a band-pass signal.

A bandpass signal can be represented as a modulated form of lowpass signals. The lowpass signals are, effectively, upconverted to bandpass signals. If we have a bandpass noise signal n(t), it can be represented in terms of lowpass signals n c (t) and n s (t):

The component n c (t) is called the inphase component, and the component n s (t) is called the quadrature component. Both are lowpass signals that are upconverted by cos  c t and sin  c t respectively. Now, let us relate the lowpass signals with the quadrature representation.

Taking the Fourier transform of the quadrature representation of n(t), we have

Let n c (t) and n s (t) have spectra which look like f N c (f), N s (f) The spectrum of n(t) will look like what we see on the following slide. B A

f Re{N(f)} f Im{N(f)} A/2 fcfc -f c

The magnitude of the spectrum of N(f) is as follows: f |N(f)| B A/  2 (The magnitude or height can be found by taking the square-root of the sum of the squares of the real and the imaginary parts of N(f).)

The power spectral density of n(t) is proportional to the square of its spectrum. f S n (f) B A 2 /2 The power in n(t) can be calculated by integrating the power spectral density.

P = 2 [ BA 2 /2] = BA 2. If we compute the power spectral density and the power of n c (t) and n s (t) we get the same value: f S nc (f), S ns (f) A2A2 B

Thus, we can conclude that the power in each the in-phase and quadrature components, n c (t), n s (t), are the same as that of the noise itself, n(t).

Exercise: Suppose we had narrow-band noise whose power spectral density is N 0 /2. Find the power spectral density magnitude of the in-phase and quadrature-phase components.

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