Flashcards
Digital Signal Processing
Flashcards do conjunto (53)
- 1SignalSequence of numbers
- 2Signal can be1) Continuous - exists at any time point -> analog
2) Discrete - exists at specific time point -> digital - 3Digital signal should be1) Discrete - achieved by sampling
2) Quantilisation: the value should be an integer - 4Signal: - Even - Odd - None- x[-n]=x[n]
- x[-n]=-x[n]
- Example: x[n]= n^2 +n -> x[-n]=n^2-n - 5If x[n] = ev{x[n]} + odd{x[n]} : ev{x[n]} odd{x[n]}= (x[n]+x[-n])/2
= (x[n]-x[-n])/2 - 6Flippingx[n] -> x[-n]
- 7Scalingx[alpha n]
|alpha|>1 -> squezzing, downsampling
0<|alpha|<1 -> expand, upscaling -> zero-pading - 8Shiftingx[n-n0]
n0>0 -> shift to right
n0<0 -> shift to left - 9Periodic signal vs non-periodicPeriodic: - ∃T ∈ R (real numbers) x(t+T)=x(t)
-∃k ∈ Z (integers numbers) x[n+k]=x[n]
Non-periodic: one instance is enough to say
*∃T ∈ R there exists an T belonging to real number - 10Fundamental periodsin/cos(wt) -> P = 2pi/w
- 11Unit impulse signalPa(t)= 1/2a at -a<=t<=a and 0 otherwise
δ(t)= 1 at t=0 and 0 at t not equal 0
Area from -inf to inf => integral from -a to a -> 1
Limit of a -> 0:
at -a<=t<=a -> undefined
otherwise -> 0 - 12Integral of f(t)δ(t) -> Integral of f(t)δ(t-t0) ->f(0)
f(t0) - 13Unit stepu(t) = 1 at t>=0
0 at <0
δ(t) = du(t)/dt
u(t) = integral (-inf to inf) of δ(t)
Small problem:
both δ(t) & u(t) do not exist - 14System: MemoryMemoryless -> output depends on input at the same time
y[n]=Rx[n] -> memoryless
y[n] = x[-n] -> with memory - 15System: invertibleA distinct input to a distinct output
- 16System: causalityA causal system is the one in which the output y(t) at time n depends only on the current input x(t) at time t, and/or its past input sample values such as x(n1), x(n2),.... Otherwise, if a system output depends on the future input values such as x(n + 1), x(n + 2),..., the system is noncausal.
All memoryless systems - causal. - 17System: stabilityevery bounded input produces a bounded output (BIBO)
- 18System: time-invarianty1(n) is the system output for the input x1(n). Let x2(n)1=x1(n-n0) be shifted version of x1(n) by n0 samples. The output y2(n) obtained with the shifted input x2(n) = x1(n - n0) is equivalent to the output y2(n) acquired by shifting y1(n) by n0 samples, y2(n) = y1(n n0).
- 19System: linearity1. Combined response:
y of x=alphax1+betax2
2. Individual response
alpha y1+ beta y2
If equal, linear
- 20Unit impulse response
- 21From δ[n] being even signalδ[n-k]=δ[-(n-k)]=δ[k-n] => symmetrical
- 22Convolution x[n]*δ[n]Sum of k -inf to inf of x[k]δ[k-n]
at time we sum all values with k
or x[k]δ[n-k] (sifting property) - 23Convolution of 2 discrete signals1. Represent one signal as a sum of impulse responses (flipping here meaningle due to the property of impulse signal δ[n-k]=δ[-(n-k)]=δ[k-n])
2. Multiply your second signal by the sum of your first signal and solve the received sum - 24For LTI: y[n] = x[n]*h[n] =>h[n]*x[n]
- 25For LTI: y[n] = x[n]*(h1[n]+h2[n]) =>x[n]*h1[n]+x[n]*h2[n]
- 26For LTI: x[n]*(h1[n]*h2[n]) =>(x[n]*h1[n])*h2[n]
- 27For LTI: memory- memory: h[n] has a non-zero value at n not equal 0
- memoryless h[n] = kδ[n] - 28For LTI: invertiblex[n] -> h[n] - h-1[n] -> x[n]
- 29For LTI: causalityh[n]=0 for n<0
as x[k-(-n)] -> y[n] - 30For LTI: stabilityDiscrete: the impulse response summable sum of k -inf to inf of |h[n]|
- 31The sum of k 0 to inf of a^k => The sum of k -inf to inf of alpha^k =>1/(1-a), where |a|<1
1-alpha^(n+1)/ 1-alpha where n>= 0 and 0<|alpha|<1 - 32Discrete Fourier Transform Inverse Discrete Fourier Transformpi to -pi
- 33Convolution: Unit step signalU[n]->LTI->s[n]
U[n]=δ[n]-δ[n-1] / h[n]=s[n]-s[n-1]
δ(t)=dU(t)/dt / h(t)=ds(t)/dt
U[n] = sum of k 0 to inf of δ[n-k]/s[n] = sum of k 0 to inf of h[n-k]
U(t)= Int of -inf to t of δ[theta] / s(n) = Int of -inf to t of h[theta] - 34DFT: Linearityx1[n]<->X1(w) x2[n]<->X2(w)
ax1[n]+bx2[n]<->aX1(w)+bX2(w) - 35DFT: Time reversalx[n]<->X(w)
x[-n]<->X(-w) - 36DFT: Frequency symmetryIf x[n] real -> X(w)=X*(-w)
* - conjugate: a+jb -> a-jb
- 37DFT: Timeshift and frequency shift2pif=w
- 38DFT: Convolution
- 39DFT: Parseval's relationshipe^jw = w -> X(w)
Energy of signal in both time and freq domains
- 40x[n]=1 -> x[n]=e^jw0n ->
![x[n]=1 -> x[n]=e^jw0n ->](https://examtimeassets.s3.amazonaws.com/uploads/media/image/38703582/desktop_d7f55b6d-1e14-4bae-8e88-ca52546bcccc.png)
- 41Z-transformsum from -inf to inf of x[n]*z^-n
- 42ZerosValues that result in H(z) being 0
- 43PolesValues that result in non-existence of H(z)
- 44If the ROC includes the unit circle + when all the poles are inside the unit circleX(z) is stable
- 45Initial value x[0]=lim of Z->inf of X(Z)
- 46Components of digital filter
- 47Finite Impulse response- Feedforward filter
- Each output sample is an algebraic weighted sum of inputs or shifted samples
- Impulse response characterising it - 48Infinite Impulse response- Each output sample is the combination of feedforward inputs and feed backward/feedback outputs
- This is driven from analog filter - 49L point moving average filter (running sum)
- 50Poles of FIRz=0, number depends on the order of filter
- 51LowpassA filter that passes frequencies f ∈ [0,fc] where fc is the cutoff frequency
Lowpass only when 0 is included - 52HighpassA filter that passes frequencies f ∈ [fc,inf) where fc is the cutoff frequency
Lowpass only when 0 is included - 53BandpassA filter which passes 0
Informações
- Descrição
- Digital Signal Processing FlashCards sobre Digital Signal Processing, criado por Alexandra Shabanova em 01-06-2021.
- Criado por
- Alexandra Shabanova
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