Outline. Discrete time signals. Impulse sampling z-transform Frequency response Stability INF4420. Jørgen Andreas Michaelsen Spring / 37 2 / 37

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Transcription:

INF4420 Discrete time signals Jørgen Andreas Michaelsen Spring 2013 1 / 37 Outline Impulse sampling z-transform Frequency response Stability Spring 2013 Discrete time signals 2 2 / 37

Introduction More practical to do processing on sampled signals in many cases Sampled + quantized signals = digital Inputs and outputs are not sampled How does sampling affect the signals? Tools for analyzing sampled signals and systems ( discrete Laplace transform, the z-transform) Spring 2013 Discrete time signals 3 3 / 37 Introduction We have already seen sample and hold circuits We can also realize integrators, filters, etc. as sampled analog systems switched capacitor techniques. Discrete time, continuous amplitude. Digital processing is efficient and robust, usually preferred where applicable. Sampling also applies to digital. Spring 2013 Discrete time signals 4 4 / 37

Introduction Spring 2013 Discrete time signals 5 5 / 37 Introduction Sample a continuous time input signal at uniformely spaced time points. Output is a discrete sequence of values (in theory). Spring 2013 Discrete time signals 6 6 / 37

Introduction Spring 2013 Discrete time signals 7 7 / 37 Sampling Laplace transform: Input signal Fourier transform: Spring 2013 Discrete time signals 8 8 / 37

Sampling Spring 2013 Discrete time signals 9 9 / 37 Sampling Impulse sampling Choose τ infinitely narrow Choose the gain k = 1 τ. The area of the pulse at nt is equal to the instantaneous value of the input at nt, f(nt). The signal is still defined for all time, so we can use the Laplace transform for analysis. Spring 2013 Discrete time signals 10 10 / 37

Sampling Modelling the sampled output, f t We will model the sampled output in the time domain Then find an equivalent representation in the Laplace domain We will model each pulse independently and the whole signal by summing all pulses Spring 2013 Discrete time signals 11 11 / 37 Sampling Modeling a single pulse using step functions Step function: u t 1, t 0 0, t < 0 Single pulse: f nt u t nt u t nt τ Spring 2013 Discrete time signals 12 12 / 37

Sampling Sum all the pulses to get the sampled signal: f t = k f nt u t nt u t nt τ Spring 2013 Discrete time signals 13 13 / 37 Sampling Transforming the time domain model to the Laplace domain Relevant Laplace transforms f(t) F(s) u(t) s 1 f t a u t a, a 0 e as F(s) Spring 2013 Discrete time signals 14 14 / 37

Sampling Time domain model from before f t = k f nt u t nt u t nt τ Laplace domain F s = k f nt e snt s k 1 e sτ = s e s nt+τ s f nt e snt Spring 2013 Discrete time signals 15 15 / 37 Sampling and the z-transform F s = k 1 e sτ s f nt e snt Impulse sampling: k = 1 τ, τ 0 (ex 1 + x) F s f nt e snt f nt z n Spring 2013 Discrete time signals 16 16 / 37

The z-transform z e st X z x nt z n = x n z n Delay by k samples, z k X(z) Convolution in time multiplication in z Spring 2013 Discrete time signals 17 17 / 37 Frequency response Use the Laplace domain description of the sampled signal As before, substitute s = jω X jω = x nt e jωnt n= X(jω) is the Fourier transform of the impulse sampled input signal, x(t). e jx is cyclic, e jx = cos x + j sin x Spring 2013 Discrete time signals 18 18 / 37

Frequency response We go from the z-transform to the frequency response by substituting z = e jωt As we sweep ω we trace out the unit circle Spring 2013 Discrete time signals 19 19 / 37 Frequency response Rewriting to use frequency (Hz), rather than radian frequency X f = x nt e j2πfnt n= Because e jx is cyclic, f 1 = k f s + f 1, where f 1 is an arbitrary frequency, k is any integer and f s is the sampling frequency (T 1 ). Spring 2013 Discrete time signals 20 20 / 37

Frequency response The frequency spectrum repeats. We can only uniquely represent frequencies from DC to f s 2 (the Nyquist frequency). Important practical consequence: We must band limit the signal before sampling to avoid aliasing. A non-linear distortion. Spring 2013 Discrete time signals 21 21 / 37 Frequency response Spring 2013 Discrete time signals 22 22 / 37

Frequency response If the signal contains frequencies beyond f s 2, sampling results in in aliasing. Images of the signal interfere. Spring 2013 Discrete time signals 23 23 / 37 Sampling rate conversion Changing the sampling rate after sampling We come back to this when discussing oversampled converters Oversampling = sampling faster than the Nyquist frequency would indicate Upsampling is increasing the sampling rate (number of samples per unit of time) Downsampling is decreasing the sampling rate Spring 2013 Discrete time signals 24 24 / 37

Downsampling Keep every n-th sample. Downsample too much: Aliasing Spring 2013 Discrete time signals 25 25 / 37 Upsampling Insert n zero valued samples between each original sample, and low-pass filter. Requires gain to maintain the signal level. Spring 2013 Discrete time signals 26 26 / 37

Discrete time filters Analog filters use integrators, s 1, as building blocks to implement filter functions. Discrete time filters use delay, z 1. Example: Time domain: y n + 1 = bx n + ay[n] z-domain: zy z = bx z + ay(z) H z Y z X z = b z a Spring 2013 Discrete time signals 27 27 / 37 Discrete time filters Frequency response, z = e jω (ω normalized to the sampling frequency, really z = e jωt ) H e jω = b e jω a DC is z = e j0 = 1. The sampling frequency is z = e j2π = 1 (also). Sufficient to evaluate the frequency response from 0 to π due to symmetry (for real signals). Spring 2013 Discrete time signals 28 28 / 37

Stability y n + 1 = bx n + ay[n] If a > 1, the output grows without bounds. Not stable. H z = b z a In a stable system, all poles are inside the unit circle a = 1: Discrete time integrator a = 1: Oscillator Spring 2013 Discrete time signals 29 29 / 37 IIR filters y n + 1 = bx n + ay[n] is an infinite impulse response filter. Single impulse input (x 0 = 1, 0 otherwise) results in an output that decays towards zero, but (in theory) never reaches zero. If we try to characterize the filter by its impulse response, we need an infinite number of outputs to characterize it. Spring 2013 Discrete time signals 30 30 / 37

FIR filters y n = 1 3 x n + x n 1 + x n 2 Is a FIR (finite impulse response filter). H z = 1 3 2 i=0 z i FIR filters are inherently stable but require higher order (more delay elements) than IIR. Spring 2013 Discrete time signals 31 31 / 37 Bilinear transform Mapping between continuous and discrete time Design the filter as a continuous time transfer function and map it to the z-domain s = z 1 1+s, conversely, z = z+1 1 s s = 0 maps to z = 1 (DC) s = maps to z = 1 First order approximation f s 2 Spring 2013 Discrete time signals 32 32 / 37

Sample and hold We modeled impulse sampling by letting τ 0. For the sample and hold, we use the same model, but let τ T. Use this to find the transfer function of SH Spring 2013 Discrete time signals 33 33 / 37 Sample and hold F (s) = k 1 e sτ s f nt e snt Impulse sampling 1 Sample and hold: F k 1 e st (s) = s The pulse lasts for the full sampling period, T f nt e snt Spring 2013 Discrete time signals 34 34 / 37

Sample and hold The sample and hold shapes spectrum H SH s 1 e st s Frequency (magnitude) response of the SH H SH jω = T sin ωt 2 ωt 2 Spring 2013 Discrete time signals 35 35 / 37 Sample and hold Sampled signal spectrum Sample and hold sinc response, sin x x Spring 2013 Discrete time signals 36 36 / 37

References Gregorian and Temes, Analog MOS Integrated Circuits for Signal Processing, Wiley, 1986 Spring 2013 Discrete time signals 37 37 / 37