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MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.003: Signals and Systems — Spring 2004Tutorial 2Tuesday, February 17, 2004Announcements• Problem set 2 is due this Friday.• There are no tutorials this week, but we will provide this handout, and TAs will hold extra office hours during the week.Today’s Agenda• DT Convolution– Calculating the DT convolution• CT Convolution– Calculating the CT convolution• LTI System Properties– Impulse response– Commutativity, distributivity, associativity, and time shift– Effect of stability on commutativity– Causality and stability• Signal Properties vs. System Properties• Singularity Functions251 DT ConvolutionIn class, we saw how to write any DT signal as the superposition sum of scaled, shifted impulses using thesifting property of the DT unit impulse:x[n]=+∞k=−∞x[k]δ[n − k]We define the unit impulse response, h[n], of a DT system H as the output when the input is the unit impulseδ[n]. If the system H is LTI, then the output y[n] for any arbitrary input x[n]isy[n]=H {x[n]}= H+∞k=−∞x[k]δ[n − k] =+∞k=−∞x[k] H{δ[n − k]} (by linearity)=+∞k=−∞x[k]H{δ[n]}shifted by k(by time-invariance)=+∞k=−∞x[k]h[n]shifted by k(apply system H)y[n]=+∞k=−∞x[k]h[n − k] (simplify)So, the output y[n]istheconvolution of the input x[n] and the unit impulse response h[n] of the LTI system.We write this as y[n]=x[n] ∗ h[n].One useful sanity check when performing DT convolutions is that a signal of length m convolved with asignal of length n produces a signal with no more than m + n − 1 non-zero terms.1.1 Calculating the DT convolutionCalculating the convolution of two DT signals can be a rather tedious process. Consider the followingexample.Given x[n]=δ[n +1]− 2δ[n]+3δ[n − 1] and h[n]=δ[n]+4δ[n − 1] − 2δ[n − 2] + δ[n − 3], calculatey[n]=x[n] ∗ h[n]. We can calculate the convolution by applying the convolution sum directlyy[n]=∞k=−∞x[k]h[n − k]26Since x[n] is nonzero for n ∈{−1, 0, 1},y[n]=1k=−1x[k]h[n − k]y[−1] = x[−1]h[0] = 1y[0] = x[−1]h[1] + x[0]h[0] = 4 + −2=2y[1] = x[−1]h[2] + x[0]h[1] + x[1]h[0] = −2+−8+3=−7y[2] = x[−1]h[3] + x[0]h[2] + x[1]h[1] = 1 + 4 + 12 = 17y[3] = x[0]h[3] + x[1]h[2] = −2+−6=−8y[4] = x[1]h[3] = 3Thus, y[n]=δ[n +1]+2δ[n] − 7δ[n − 1] + 17δ[n − 2] − 8δ[n − 3] + 3δ[n − 4].It is sometimes quicker, and less error prone, to use the following trick when calculating the convolutionof two finite length DT signals. First we form the following table:x[n]h[n] 1 4 −211 14−21−2 −2 −84−23 312−63The table consists of the values of x[n] in the first column and the values of h[n] in the top row. The elementsin the interior of the table are formed by multiplying the corresponding elements in the first column andfirst row. Note that the underlined entries correspond to the element at zero (i.e. x[0] and h[0]). Thedouble-underlined element corresponds to the entry formed by multiplying the two underlined elements(i.e. x[0]h[0]). Now we can simply form y[n] by summing diagonally. Using sequence notation:y[n]=1, 4+(−2), (−2) + (−8) + 3, 1+4+12, (−2) + (−6), 3= {1, 2, −7, 17, −8, 3}Note again that we have underlined the element resulting from a sum containing the double-underlinedelement, which corresponds to y[0]. This result for y[n] is equivalent to the one above.27Problem 2.1Evaluate the following discrete-time convolution sums given below:(a) y[n] = cos12πn∗12nu[n − 2](b) y[n]=u[n] ∗∞p=0δ[n − 2p](c) y[n]=x[n] ∗ h[n], where x[n]andh[n] are given below:43210-1-21212112h[n]x[n]-1nn -4-3-2-1021-1-2-5rrrrrrrrrrrrrrr(Workspace)282 CT ConvolutionSimilarly, we can express CT signals as the superposition sum of scaled, shifted impulses using the siftingproperty of the CT unit impulse:x(t)=+∞−∞x(τ)δ(t − τ )dτWe likewise define the unit impulse response, h(t), of a CT system H as the output when the input is theunit impulse δ(t). If the system H is LTI, then the output y[n] for any arbitrary input x[n]isy(t)=+∞−∞x(t)h(τ − t)dτory(t)=x(t) ∗ h(t)One useful sanity check when performing CT convolutions is that a signal of length t1convolved with asignal of length t2produces a signal of length no more than t1+ t2.2.1 Calculating the CT convolutionTo graphically compute the convolution integral x(t) ∗ h(t), use the following steps:1. Plot x(τ )andh(τ ).2. Select which signal to flip, say h(τ). In general, the signal of shorter duration is the one you shouldflip to make the computation easier.3. Flip h(τ ) about the vertical axis, which gives us h(−τ ).4. Plot h(−τ + t)ontheτ axis. This is just h(−τ) shifted to the RIGHT by t. Note that in order toshift h(−τ )totherightbyt, we replace τ by τ − t which gives us h(−(τ − t)) = h(−τ + t).5. Identify different regions of overlap as t is varied from −∞ to +∞ where there are “breaks” in ei-ther function. These correspond to discontinuities and regions where the mathematical expressionsdescribing the functions change.6. Multiply both functions and integrate them for each of the regions identified above. Make sure to usethe correct limits for the integrals.29Problem 2.2 Evaluate the following continuous-time convolution integrals given below:(a) y(t)=u(t +1)∗ u(t − 2)(b) y(t)=(u(t +2)− u(t − 1)) ∗ u(−t +2)(c) y(t)=x(t) ∗ h(t), where x(t)andh(t) are given below:12t2-212-1 0 1x(t) y(t)0t1-12-3-2(Workspace)303 LTI System Properties3.1 Impulse responseThe impulse response completely characterizes an LTI system, i.e. there is a one-to-one correspondencebetween the set of impulse responses and the set of LTI systems. In fact, we can even define or identify anLTI systems by its impulse response, h(t). If we want to find h(t), we can set the input x(t)=δ(t):y(t)=h(t) ∗ δ(t)= h(t)For example, to find the impulse response, h(t), of the systemy(t)=−x(t − t2)+6t−∞x(τ − t1)dτ,we substitute x(t)=δ(t):h(t)=−δ(t − t2)+6t−∞δ(τ − t1)dτ=6u(t − t1) − δ(t − t2)31Problem 2.3 Find the impulse response of the following systems:(a) y[n]=2x[n]+3x[n − 1] − 4x[n − 2](b) y[n]=nk=−∞x[k](c) y(t)=t−∞x(τ − 7)dτ(Workspace)323.2 Commutativity, distributivity, associativity, and time shiftThe convolution


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