EECE 290 Homework 9, Spring 2015

Homework 9 P20.1.4 Evaluate the following convolution operations: (a) (b) (c) (Note that sinhx = jsinjx). Solution: (a) sintcost  = . (b) costcost  cost + . (c) sinhat * sint      . P20.1.9 The switch in Figure P20.1.9 is moved at t = 0 from position ‘a’ to position ‘b’ after being in position ‘a’ for a long time. Determine i for t  0, assuming V. Solution: The response to vSRC will be determined first, assuming zero initial conditions, and then the initial conditions added by superposition. To determine the response to vSRC by convolution, it is necessary to determine the impulse response. The circuit for determining the response to a unit impulse is shown. In response to the impulse, the – + 2 H 1 H t = 0 5 A 1  6  vSRC P20.1.9 i a b – + 3H 6  (t) i , s 8 20 8.1 7.6 0.4 inductor behaves as an open circuit, the impulse appears across the inductor, resulting in a current step of 1/3 A;  = 3/6 = 0.5 s. It follows that A. Using the convolution integral, = . Initially, the current in the 1 H inductor is zero, and that in the 2 H inductor is 5 A. At t = 0+, the current in the two inductors is equalized in accordance with conservation of flux linkage. At t = 0-, the flux linkage is 10 Wb- T; at t = 0+, i(0+) = 10/3 A, and . The total current is: A. P20.1.10 An input defined as: vI = (t +1) V, -1  t  0, vI = (-t +1) V, 0  t  1, and vI = 0 elsewhere, is applied to a circuit having the impulse response . Determine the output for all t. Solution: For . For = . P20.1.12 The response of a linear, time-invariant circuit to a unit step input, u(t), is . Determine the response to an input . Solution: The response h(t) to a unit impulse the time derivative of the response to a unit step, that is . The required response is: = , for t  0-, or . P20.2.1 Given h(t) and x(t) in Figure P20.2.1. Determine h(t)*x(t) at t = 8.1 s. Solution: When h(t) is folded around the vertical axis and shifted by 8.1 s to the right, there will be an overlap of 0.4 s, as indicated by the shaded area. This area is 20200.4 = 160. P20.2.2 Given f(t) in Figure P20.2.2. Determine f(t)*2u(t) at t = 5 s. Solution: f(t)*2u(t) = twice the area of f(t) = area of the rectangle 34 = 12. P20.2.3 Given h(t) and x(t) in Figure P20.2.3. Determine h(t)*x(t) at t = 6 s. Solution: When x(t) is folded around the vertical axis and shifted by 6 s to the right, the t, s 0.5 20 t, s 8 20 P20.2.1 h(t) x(t) Figure P20.2.3 h(t) t, s 12 18 0 6 x(t) t, s 10 10 0 Figure P20.2.2 t, s 4 3 0 2 f(t) h() 18 10 functions will be as shown. h() = 3; x(-) = -( – 6). The convolution integral becomes: 108. P20.2.4 Given the function and f(t) in Figure P20.2.4. Determine h(t)*x(t) at t = 3 s. Verify the result by direct integration. Solution: Method 1 – Graphical Solution : If the pulse is folded around the vertical axis and shifted to the right by 3 s to the right, the figure becomes a shown. The convolution integral, shown shaded, is: . Method 2 – Graphical Solution : If the exponential function is folded around the vertical axis and shifted to the right by 3 s, it becomes as shown. The exponential function becomes . The convolution integral, shown shaded, is: , as before. Method 3 – Graphical Solution : f() can be expressed as the sum of two step functions: 2u() – 2u( – 2). Convolving with the step 2u() gives: . Convolving with the step -2u( – 2) involves negating the function and replacing t by (t – 2), that is, the convolution integral becomes . Adding the two integrals gives: . At t = 3 s, the result is the same as before. Method 4 – Analytical Solution : f() = 2u() – 2u( – 2); x(t – ) = . The first integral is: . The second integral is: = . Adding the two integrals gives: . At t = 3 s, the result is the same as before. t, s 2 P20.2.4 f(t) 2 1 1