According to the ACR Manual of MR Safety, Level 2 MR Technol…
According to the ACR Manual of MR Safety, Level 2 MR Technologists performing human scanning should always be considered the primary emergency responder for other Level 2 MR Technologists conducting MR examinations simultaneously.
Read DetailsWhich level of MR Personnel are those who have been educated…
Which level of MR Personnel are those who have been educated and successfully mastered MR safety topics as defined by the facility’s MRMD to ensure that they would not constitute a danger to themselves or others in the MR environment?
Read DetailsProblem 1 (8 points) The signal \( s(t) = (1 – e^{-2t})u(t)…
Problem 1 (8 points) The signal \( s(t) = (1 – e^{-2t})u(t) \) is the step response of a linear time-invariant (LTI) system. Which of the following is the output \( y(t) \) when the input is \( x(t) = 2e^{-2t}u(t) \)? (a) \( 4(1 – e^{-2t})u(t) \)(b) \( 4te^{-2t}u(t) \)(c) \( (1 – e^{-2t})u(t) \)(d) \( 4te^{-2t}u(t) \)(e) \( (4 – e^{-2t})u(t) \)(f) \( (1 – 2e^{-4t})u(t) \)(g) \( 4tu(t) \)(h) \( 4e^{-4t}u(t) \) Problem 2 (8 points) Find the expression of \(i(t)\) (in unit A) for all times \(t\) in the figure below. (a) \(i(t) = \frac{1}{2}\left(1-\mathrm{e}^{-1200t} \right)u(t)\)(b) \(i(t) = \frac{1}{2}\left(1-\mathrm{e}^{-800t} \right)u(t)\)(c) \(i(t) = \frac{1}{2}\left(1-\mathrm{e}^{-2400t} \right)u(t)\)(d) \(i(t) = \frac{1}{2}\left(1-\mathrm{e}^{-3600t} \right)u(t)\)(e) \(i(t) = \frac{3}{4}\left(1-\mathrm{e}^{-1200t} \right)u(t)\)(f) \(i(t) = \frac{3}{4}\left(1-\mathrm{e}^{-800t} \right)u(t)\)(g) \(i(t) = \frac{3}{4}\left(1-\mathrm{e}^{-2400t} \right)u(t)\)(h) \(i(t) = \frac{3}{4}\left(1-\mathrm{e}^{-3600t} \right)u(t)\) Problem 3 (8 points) Obtain the current \(i(t)\) (in unit A) for all values of time \(t\), in the circuit of the figure below. (a) \(i(t) = \frac{7}{2}\mathrm{e}^{-25t}u(t)+1\)(b) \(i(t) = \frac{7}{2}\mathrm{e}^{-100t}u(t)+1\)(c) \(i(t) = \frac{9}{2}\mathrm{e}^{-25t}u(t)+2\)(d) \(i(t) = \frac{9}{2}\mathrm{e}^{-100t}u(t)+2\)(e) \(i(t) = -\frac{7}{2}(1-\mathrm{e}^{-25t})u(t)+1\)(f) \(i(t) = -\frac{7}{2}(1-\mathrm{e}^{-100t})u(t)+1\)(g) \(i(t) = -\frac{9}{2}(1-\mathrm{e}^{25t})u(t)+2\)(h) \(i(t) = -\frac{9}{2}(1-\mathrm{e}^{100t})u(t)+2\) Problem 4 (8 points) Derive the impulse response \(h(t)\) of the voltage \(v(t)\) for RL parallel circuit in figure below, given the impulse input \(\delta(t)\) of the current source \(i_s(t)\). (a) \(\mathrm{e}^{-\frac{R}{L}t} u(t)\) (b) \(\frac{1}{R} \left(1-\mathrm{e}^{-\frac{R}{L}t}\right) u(t)\) (c) \(R\mathrm{e}^{-\frac{R}{L}t}u(t)\) (d) \(\left(1-\mathrm{e}^{-\frac{R}{L}t}\right)u(t)\) (e) \(\frac{R}{L} \mathrm{e}^{-\frac{R}{L}t} u(t) + \delta(t)\) (f) \(-\frac{1}{L} \mathrm{e}^{-\frac{R}{L}t} u(t)\) (g) \(- \frac{R^2}{L} \mathrm{e}^{-\frac{R}{L}t} u(t) + R \delta(t)\) (h) \(\frac{R}{L}\mathrm{e}^{-\frac{R}{L}t}u(t)\) Problem 5 (8 points) Suppose we have an LIT system with impulse response {“version”:”1.1″,”math”:””}\(h(n)=\delta(n-1)-3\delta(n-3)\), an unknown input \(x(n)\) results output {“version”:”1.1″,”math”:””}\(y(n)=sin(2t-2)-3sin(2t-6)+e^{-4t+4}-3e^{-4t+12}\), find the unknown input \(x(t)\) (a) \(x(t)=sin(2t-2)-3sin(2t-6)+e^{-4t+4}-3e^{-4t+12}\)(b) \(x(t)=sin(2t-3)-3sin(2t-9)+e^{-4t+3}-3e^{-4t+9}\)(c) \(x(t)=sin(2t-1)-3sin(2t-3)+e^{-4t+3}-3e^{-4t+15}\)(d) \(x(t)= sin(2t)+e^{-4t}\)(e) \(x(t)= sin(2t)-3e^{-4t} \)(f) \(x(t)= sin(2t)+3e^{-4t} \)(g) \(x(t)= 3sin(2t)+3e^{-4t} \)(h) \(x(t)= 3sin(2t)-3e^{-4t} \) Problem 6 (8 points) Find the inverse Laplace Transform of the following: $$F(s) = \frac{6s+22}{(s+3)^2+4}$$(a) \(f(t)=6e^{-3t} +2e^{-3t}{s+5}\)(b) \(f(t)=2e^{-3t}cos(2t)+5e^{-3t}sin(2t)\)(c) \(f(t)=3e^{-6t}cos(2t)-2e^{-6t}sin(2t)\)(d) \(f(t)=6e^{-3t}cos(2t)+2e^{-3t}sin(2t)\)(e) \(f(t)=cos(2t)+sin(2t)\)(f) \(f(t)=6e^{-2t}sin(3t)+3e^{-2t}cos(3t)\)(g) \(f(t)= 2e^{-3t}sin(2t)\)(h) \(f(t) = 3e^{-2t}cos(3t)\) Problem 7 (8 points) Which of the following is the Laplace transform of \(x(t)=t\,e^{-3t}u(t)\)?(a) \(\displaystyle \frac{1}{(s+3)^2}\)(b) \(\displaystyle \frac{s}{(s+3)^2}\)(c) \(\displaystyle \frac{1}{s+3}\)(d) \(\displaystyle \frac{1}{s+3} – \frac{3}{(s+3)^2}\)(e) \(\displaystyle \frac{2}{(s+3)^2}\)(f) \(\displaystyle \frac{1}{(s+3)^3}\)(g) \(\displaystyle \frac{s+3}{(s+3)^2}\)(h) \(\displaystyle \frac{3}{(s+3)^3}\) Problem 8 (8 points) A causal LTI system has impulse response \[h(t)=\bigl(1-e^{-3t}\bigr)u(t)\]What input \(x(t)\) yields the unit‑step output \(y(t)=u(t)\)? (a) \(x(t)=\delta(t)-\delta(t-3)\)(b) \(x(t)=\tfrac{1}{3}\,\delta(t)+u(t)\)(c) \(x(t)=3\,e^{-3t}u(t)\)(d) \(x(t)=3\,u(t)\)(e) \(x(t)=\delta(t)+3e^{-3t}u(t)\)(f) \(x(t)=3\delta(t)-3e^{-3t}u(t)\)(g) \(x(t)=\delta(t)\)(h) \(x(t)=\delta(t)+\delta(t-3)\) Problem 9 (18 points) The convolution of two signals is defined by{“version”:”1.1″,”math”:””}\[f(t) = 3[u(t + 3) – u(t – 3)], \hspace{6 mm}g(t) = 2[u(t + 2) – u(t – 2)].\]Let \( h(t) = f(t) * g(t) \). Draw the resulting signal \( h(t) \). Clearly label: The time axis and amplitude axis The start and end times of the signal The value of the maximum amplitude The coordinates of all corner points in the sketch Problem 10 (18 points) Given the finite–length sequences g [ n ] = [ 1 0 2 3 1 4 ] , f [ n ] = [ 2 1 0 1 ] , {“version”:”1.1″,”math”:”\[ g[n] = [\,1\;\; 0\;\; 2\;\; 3\;\; 1\;\; 4\,], \qquad f[n] = [\,2\;\; 1\;\; 0\;\; 1\,], \] “}compute the discrete linear convolution \(y[n] = g[n] * f[n]\) and write all non-zero values of \(y[n]\). 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