Fundamentals of Noise V.Vasudevan, Department of Electrical Engineering, Indian Institute of Technology Madras
Noise in resistors Random voltage fluctuations across a resistor Mean square value in a frequency range f proportional to R and T Independent of the material, size and shape of the conductor Contains equal (mean square) amplitudes at all frequencies (upto a very high frequency) Power contained in a frequency range f is the same at all frequencies or the power spectral density is a constant.
Questions Why do these fluctuations occur? Electrons have thermal energy a finite velocity At random points in time, they experience collisions with lattice ions velocity changes Velocity fluctuations lead to current fluctuations What does the power spectral density of a random signal mean?
Background Background on random variables A sample space of outcomes that occur with certain probability A function that maps elements of the sample space onto the real line - random variable Discrete random variables - example is the outcome of a coin tossing experiment Continuous random variables - Voltage across a resistor
Background Continuous Random variables 0.4 Continuous random variables are characterized by probability density functions (pdfs). Examples are 1 1.4 1.2 0.8 0.3 1 0.6 N(0,1) 0.2 f(x) 0.8 f(x) 0.6 0.4 0.1 0.4 0.2 0.2 0-6 -4-2 0 2 4 6 x 0-1 -0.5 0 0.5 1 x 0 0 2 4 6 8 10 x Gaussian Uniform Exponential f (x)dx = 1
Background Moments of random variables Mean Variance Correlation Covariance µ x = E{X } = R XY = E{XY } = σ 2 = E{X 2 } µ 2 x xf (x)dx xyf XY (x, y)dxdy K XY = E{(X µ X )(Y µ Y )} If K XY = 0, X and Y are uncorrelated. The correlation coefficient is c = K XY σ X σ Y.
Random Processes Random Processes Random variable as a function of time The value at any point in time determined by the pdf at that time The pdf and hence the moments could be a function of time If {X 1,.....X n } and {X 1 + h,....x n + h} have the same joint distributions for all t 1,... t n and h > 0, it is n th order stationary We will work with wide sense stationary (WSS) processes - The mean is constant and the autocorrelation, R x (t, t + τ) = R x (τ)
Random Processes Spectrum Finite energy signals - eg. y(t) = e at. The Fourier transform exists and energy spectral density is Y (f ) 2 Finite power signals - eg. y(t) = sin(2πf o t) + sin(6πf o t). In this case, can find the spectrum of the average power - eg. for y(t) it is 1 2 at frequencies f o and 3f o. Random signals are finite power signals, but are not periodic and the Fourier transform does not exist. The question is how do we find the frequency distribution of average power?
Random Processes Limit the extent of the signal x T (t) = x(t), T /2 t T /2 Find the Fourier transform X T (f ) of this signal. The energy spectral density is defined as ESD T = E{ X T (f ) 2 } The power spectral density (PSD) is defined as The total power in the signal is E{ X T (f ) 2 } S x (f ) = lim T T P x (f ) = S x (f )df
Random Processes Weiner-Khinchin Theorem The inverse Fourier transform of X T (f ) 2 is x T (t) x T ( t) i.e. [ F 1 E{XT (f ) 2 ] } = 1 T /2 E{x T (t)x T (t + τ)}dt T T T /2 But for WSS signals, R X (τ) = E{x T (t)x T (t + τ)} Therefore, F [R x (τ)] = S x (f )
Random Processes Noise in resistors Simple model Collisions occur randomly - Poisson process Velocities before and after a collision are uncorrelated Average energy per degree of freedom is kt 2 With these assumptions it is possible to show that S I (f ) = 4kT R the autocorrelation is a delta function (white noise). Also, using central limit theorem, it has a Gaussian distribution. However, continuous time white noise has infinite power and is therefore an idealization.
Random Processes Representation of a noisy resistor S v (f) = 4kTR S I (f) = 4kT R
LTI systems Noise in Linear time-invariant systems If h(t) is the impulse response of an LTI system, y(t) = h(t) x(t) Assume x(t) is a WSS process with autocorrelation R x (τ) E{y(t)y(t + τ)} = = h(α)h(β)e{x(t α)x(t + τ β)}dαdβ h(α)h(β)r x (τ + β α)dαdβ y(t) is also a WSS process and R y (τ) = h(τ) h( τ) R x (τ)
LTI systems PSD and Variance If H(f ) = F{h(t)}, the power spectral density at the output is S y (f ) = F{R y (τ)} = H(f ) 2 S x (f ) Noise power at the output is R y (0) = S y (f )df In linear systems, if the input has a Gaussian distribution, the output is also Gaussian
LTI systems Example - RC Circuit The output PSD is S=4kTR R C V o S y (f ) = 4kTR 1 + 4π 2 f 2 R 2 C 2 The total power at the output is P y = 0 4kTR 1 + 4π 2 f 2 R 2 C 2 df = kt C
LTI Two port Linear two ports - RLC circuits Resistors are the noise sources Assuming N noise sources, the in terms of the y parameters, the output current can be written as I in = y ii V in + y io V o + I out = y oi V in + y oo V o + N y ij V j j=1 N y oj V j j=1 V j is the random noise source due to the j th resistor
LTI Two port Noisy RLC two ports - Representation If I ni = N j=1 y ijv j and I no = N j=1 y ojv j, the two-port can be represented as a noiseless network with two additional noise current at the two ports. Can also have a Z network, with noise voltage sources at the two ports v ni v no + + I ni Y I no Z To get I ni and I no, find the short circuit current at the two ports. For v no and v ni, find the open circuit voltage
LTI Two port Examples R 1 R 3 R 2 R 1 R 2 v 2 ni = 4kT (R 1 + R 2 ) f v 2 no = 4kT (R 3 + R 2 ) f The two noise sources are correlated I 2 ni = 4kT R 1 f I 2 no = 4kT R 2 f The two sources are uncorrelated
LTI Two port Generalized Nyquist theorem Supposing we have an RLC network and we wish to find the noise at the output port. First remove all noise sources. Connect a unit current source I o at the output Voltage across resistor R j due to I o is V j = Z jo Power absorbed in R j is P j = Z jo 2 in the network is P = R j N Z jo 2 j=1 Power delivered to the network by the source is R j P = Re(V o I o ) = Re(Z o ) Total power dissipated
LTI Two port Power delivered = Power dissipated N Z jo 2 Re(Z o ) = Now include noise current sources due to resistors. Output voltage due to noise generated by resistors is j=1 R j V o 2 = 4kT f N j=1 Z oj 2 R j Since the network is reciprocal Z oj = Z jo V o 2 = 4kTRe(Zo ) f Called the Generalized Nyquist theorem
LTI Two port V n I n Representation v n + i n Replace all noise sources in the network by an equivalent noise voltage and current source in the input port, so that the correct output noise spectral density is obtained Once again, the two sources could be correlated
LTI Two port Comparing with the I n I n representation, The correlation coefficient is V n = I no y oi I n = I ni y ii y oi I no c iv (f ) = E{I n V n } [E{I 2 n }E{V 2 n }] 1 2 S iv (f ) = [S ni (f )S nv (f )] 1 2
V n+ Fundamentals of Noise LTI Two port Noise figure of a two-port network Y s I ns I n F = Total noise power at the output per unit bandwidth Output noise power per unit bandwidth due to the input F = E{ I ns + I n + Y s V n 2 } E{ I ns 2 } = 1 + S ni(f ) S s (f ) + Y s 2 S nv(f ) 2 + 2Re(c iv Ys ) [S ni(f )S nv (f )] 1 2 S s (f ) S s (f ) Depending on the correlation coefficient, one can use the right type of source impedence to minimize noise figure
LTI Two port References 1. H.A.Haus et al, Representation of noise in linear twoports, Proc. IRE, vol.48, pp 69-74, 1960 2. H.T.Friss, Noise figure of radio receivers, Proc. IRE, vol.32, 1944, pp 4-9-423