What is ? Fluctuations in either the imaging system or the object being imaged. Bioengineering 280A Quantization Noise: Due to conversion from analog waveform to digital number. Principles of Biomedical Imaging Quantum Noise: Random fluctuation in the number of photons emittted and recorded. Fall Quarter 2008 MRI Lecture 6: Noise and SNR Thermal Noise: Random fluctuations present in all electronic systems. Also, sample noise in MRI Other types: flicker, burst, avalanche - observed in semiconductor devices. Structured Noise: physiological sources, interference

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

Quantization Noise Physiological Noise

Perfusion Time Series

Signal s(t) Perfusion time series: Before Correction (cc = 0.15)

Quantized Signal r(t) Perfusion time series: After Correction (cc = 0.71) Signal Intensity Quantization noise Image Number (4 Hz) q(t) Pulse Ox (finger) r(t) = s(t) + q(t) Cardiac component estimated in brain Although the noise is deterministic, it is useful to model the noise as a random process. Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

1 Noise and Image Quality Thermal Noise Fluctuations in voltage across a resistor due to random thermal motion of electrons. Described by J.B. Johnson in 1927 (therefore sometimes called Johnson noise). Explained by H. Nyquist in 1928.

V 2 = 4kT " R" BW

Bandwidth Variance in Voltage Resistance

! Temperature

Prince and Links 2005 Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

Thermal Noise Thermal Noise

V 2 = 4kT " R" BW Noise is independent of up 13 At room temperature, noise in a 1 k# resistor is to 10 Hz. Therefore it is a source of . V 2 / BW =16 $10%18 V 2 / Hz Amplitude distribution of the noise is Gaussian. In root mean squared form, this corresponds to V/BW = 4 nV/ Hz . ! Example : For BW = 250 kHz and 2 k# resistor, total noise voltage is 2 "16 $10-18 "250 $103 = 4 µV

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008 !

2 Signal in MRI Signal in MRI Recall the signal equation has the form Signal in the receiver coil t " t /T2 ( r) " j#0t ( + t # t /T2 ( r) # j"0t ( + sr (t) = M(x,y,z)e e exp* " j$ G % ' r(% )d% -dxdydz & & & &0 ( ) sr (t) = j"0B1xy M(x,y,z)e e exp* # j$ G % ' r(% )d% -dV ) , & ) &0 ( ) ,

Recall, total magnetization is proportional to B0

B0 ! Also "0 = $B0 . Therefore, total signal is proportional to B2 Faraday's Law 0 #$ EMF = " #t Magnetic Flux = B (x,y,z) M(x,y,z)dV ! $ = & 1 %

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

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Noise in MRI SNR in MRI Primary sources of noise are: signal amplitude B2 1) Thermal noise of the receiver coil SNR = " 0 2) Thermal noise of the sample. standard deviation of noise 1/ 2 2 #B0 + $B0 Coil Resistance: At higher , the EM waves tend to travel along the surface of the conductor (skin If coil noise dominates effect). As a result, R 1/2 N 2 1/2 B1/ 2 coil " #0 $ coil " #0 " 0 7 / 4 SNR " B0 Sample Noise : Noise is white, but differentiation process due to Faraday's law introduces a multiplication If sample noise dominates by #0 . As a result, the noise variance from the sample 2 is proportional to #0 .

SNR " B0 2 2 2 Nsample "#0 " B0

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

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! 3 Random Variables Mean and Variance A random variable X is characterized by its cumulative

distribution function (CDF) µX = E[X] # Pr(X " x) = FX (x) = xf (x)dx $"# X The derivative of the CDF is the probability density

function(pdf) 2 % X = Var[X]

fX (x) = dFX (x) / dx 2 = E[(X " µX ) ] The probability that X will take on values between two # limits x and x is = (x " µ )2 f (x)dx 1 2 $"# X X 2 2 x2 Pr(x " X " x ) = F (x ) # F (x ) = f (x)dx = E[X ]" µX 1 2 X 2 X 1 $x X 1

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

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Gaussian Random Variable Independent Random Variables f (x ,x ) f (x ) f (x ) X1 ,X 2 1 2 = X1 1 X 2 2

E[X1X 2 ] = E[X1]E[X 2] 1 f (x) = exp $(x $ µ)2 / 2# 2 X 2 ( ( )) Let Y = X X then 2"# 1 + 2 µY = E[Y]

= E[X1]+ E[X 2 ] µX = µ = µ1 + µ2 2 2 2 2 2 2 2 # X = # E[Y ] = E[X1 ]+ 2E[X1]E[X 2]+ E[X 2 ] = E[X1 ]+ 2µ1µ2 + E[X 2 ] 2 2 2 "Y = E[Y ]# µY 2 2 2 2 = E[X1 ]+ 2µ1µ2 + E[X 2 ]# µ1 # µ2 # 2µ1µ2 = " 2 + " 2 ! X1 X 2

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008 !

4 Signal Averaging We can improve SNR by averaging. Noise in k-space Let Recall that in MRI we acquire samples in k -space. y1 = y0 + n1 y = y + n The noise in these samples is typically well described 2 0 2 by an iid random process. For Cartesian sampling, the noise in the image domain The sum of the two measurements is 2y + n + n . 0 ( 1 2 ) is then also described by an iid random process. If the noise in the measurements is independent, then S(k) 2 the variances sum and the total variance is 2" n For each point in k -space, SNR = where " n S(k) is the signal and " n is the standard deviation of 2y0 ! ! SNRTot = = 2SNRoriginal each noise sample. 2" n

In general, SNR # Nave # Time

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

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Noise in image space Signal Averaging

2 Noise variance per sample in k - space is " n . Each voxel in image space is obtained from the Fourier transform We can improve SNR by averaging in k - space of k - space data. Say there are N points in k - space. The overall noise variance In general, SNR " Nave " Time 2 contribution of these N points is N" n . If we assume a point object, then all points in k - space contribute

equally to the signal, so overall signal is NS0. Then overall SNR in image space is ! NS S SNR # 0 = N 0! ! N" n " n Therefore, SNR increases as we increase the matrix size.

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

! 5 Effect of Readout Window SNR and Phase Encodes ADC samples acquired with sampling period "t. Assume that spatial resolution is held constant. 2 1 Thermal noise per sample is # n $"f = What happens if we increase the number of phase "t 1 encodes? Recall that "y = . Thus, increasing Wk If we double length of the readout window, the y the number of phase encodes NPE , decreases #ky and

noise variance per sample decreases by two. increases FOVy .

If we double the number of phase encodes, each point The noise standard deviation decreases by 2, and in image space has double the number of k -space lines contributing to its signal. The noise variances sum. the SNR increases by! 2. ! The SNR therefore goes up by 2.

In general, SNR T N t $ Read = kx " In general SNR$ NPE

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008

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!

Overall SNR Example

Signal $x$y$z SNR" " # n # n Putting everything together, we find that

SNR" Nave Nx NPE $t$x$y$z

= Measurement Time %Voxel Volume Assume same readout gradients for both sequences. Sampling rate for sequence 2 is twice that of sequence 1. In general, ! What are the relative! SNRs?

SNR" Measurement Time %Voxel Volume % f (&,T1 ,T2 )

Thomas Liu, BE280A, UCSD, Fall 2008 Thomas Liu, BE280A, UCSD, Fall 2008 Nishimura 1996 !

6 Example

Sampling rate for sequence 2 is twice as large, so that bandwidth is doubled. Therefore noise variance is also doubled

256A 256A SNR1 = = 2 256" n " n 512A 512A 256A SNR1 = = = 2 2" " 512(2" n ) n n

Note that sequences have! the same resolution, but sequence 2 has twice the FOV.

! Thomas Liu, BE280A, UCSD, Fall 2008

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