summary basics

7/29/2019 Summary Basics

1/12

Brief Notes on 1st Part of ME579

Would like to see X(f), the Fourier Transform of x(t),or

ck, the Fourier Series coefficients, if the time history isperiodic.

We have:x(n) samples of x(t) every seconds forn=0,1,2,3.N-1.

And can do Discrete Fourier Transforms (DFT) to give:Xk for k=0,1,2,.N-1 corresponding to frequenciesfk = 0, fs/N, 2fs/N, .(N-1)fs/N.


2/12


ISSUES: We have done three things

WINDOWEDWINDOWED(TRUNCATION - LEAKAGE)

xw(t) = w(t) . x(t)Xw(f) = W(f) convolved with X(f)

SAMPLED IN TIMESAMPLED IN TIME(ALIASING, CLIPPING, QUANTIZATION NOISE)

xS(t) = (t-n) . x(t)Periodic in Frequency.


3/12


SAMPLED IN FREQUENCYSAMPLED IN FREQUENCY

Xk = XS(f) evaluated at f= fkfk = 0, fs/N, 2fs/N, .(N-1)fs/N.

Signal now becomes Periodic in Time

xn+qN = xn


4/12

Domains: Time and Frequency

Sampled/Discrete Periodic

Multiplication Convolution

Real and Even Real and Even

Real and Odd Imaginary and Odd

Narrow Broad

Length (T or fs) Resolution (1/T or )


5/12

When do we Zero Pad Signals?

1. When result of an operation should yield alonger signal than original signal(s).e.g. convolution and time-delay.

2. When we want to have a clearer picture ofXs(f), the Fourier Transform of the sampledsignal, xs(t).

[Would prefer to transform more data to getbetter resolution, i.e., a spikier W(f).Use zero padding when we dont have any moredata to transform.]


6/12

Discrete Fourier Transform (DFT), XkRelationship to X(f) and ck

Assume no aliasing when sampling [fs > 2 fmax]Have x(n) for n=0,1,2,N-1;

Xk = D.F.T.(x); k=0,1,2,3,N-1.

Periodic Signals:N = a whole number of periods = q Tp

ck = Xk/N for -(N/2) < k < (N/2)

Transients (some aliasing will occur):X(f) X

kfor -fs/2 < f < fs/2


7/12

Discrete Fourier Transform (DFT), Xk

Symmetric about 0 and fs/2.Real Part:even symmetry about these pointsImaginary Part:odd symmetry about these points

PeriodicXk for k=+(N/2),+(N/2)+1, . N-1

equal toXk for k=-(N/2),-(N/2)+1, . 1

[fftshift will rearrange for you; you have tomake the corresponding frequency vector.]


8/12

Analog to Digital - Digital to Analog

Input/Output Range, No. of Bits, fs

Analog to Digital Conversion (ADC)

Quantization Error, Clipping, Sample and hold,

Aliasing, Anti-aliasing filters, Sample rate.

f1 = highest frequency of interest

fc = filter cut-off frequencyfmax = highest frequency in filtered signal

fs > 2 fmax = sample rate.

f1 < fc < fmax < fs/2


9/12

Analog to Digital - Digital to Analog

Digital to Analog Conversion

Similar issues as for ADC

Zero-order hold characteristicssinc function in frequency,

distorts signal in range -fmax> fmax.

Reconstruction Filter: fmax < fc


10/12

Other Things

We use the delta function,

(t) or

(f),in a lot of our theory and proofs. Sifting property in integrals Integral from - to + of exp(j2ft)

Sampling theory

We looked at the FFT algorithm (not on

exam).

Convolution of continuous and discrete

signals.


11/12

All the Transforms: timefrequency

Complex Fourier Series x(t) ck

periodic in time, discrete in frequency

Fourier Transforms x(t) X(f)

continuous in time and frequency

Fourier Transform of a sampled signal:

xs(t) Xs(f) OR x(n) Xs(f)

Xs(f) = (1/) q

X(f - q fs)

discrete in time, periodic in frequency

Discrete Fourier Transform (finite set of data used)

xn X

k; n and k: 0,1,2..N-1.

eriodic and discrete in both time and fre uenc


12/12

Using the Discrete Fourier Transform

Fourier Series coefficients(no aliasing and N corresponds to a whole no. of periods)ck = Xk/N for k=0,1,2N/2.

Approximate the Fourier Transform X(f)of x(t) for frequencies: f = k.f

s

/N(no aliasing)

X(f) at f = k.fs/N = Xk for k=0,1,2(N/2).

Sampled version of Xs(f)(sampled signal, xs(t) was of finite length = N points)

Xs(f) at f = k.fs/N = Xk

Can zero pad to evaluate at more frequency points

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