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Gambarkan output signal $y(t) = x(0.4t + 1.5)$ dari input signal berikut: $x(t) = u(t + 2) - u(t - 1)$

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Transformations of functions

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Q. Gambarkan output signal

y(t) = x(0.4t + 1.5)

dari input signal berikut:

x(t) = u(t + 2) - u(t - 1)

Understand Input Signal: First, let's understand the input signal $x(t)$ . It's a combination of two unit step functions. The first unit step function, $u(t + 2)$ , shifts the step to the left by $2$ units. The second unit step function, $u(t - 1)$ , shifts the step to the right by $1$ unit.
Apply Transformation: Now, we need to apply the given transformation $y(t) = x(0.4t + 1.5)$ to the input signal. This means we'll replace every instance of $t$ in $x(t)$ with $0.4t + 1.5$ .
Transform First Unit Step: Let's transform the first unit step function: $u(0.4t + 1.5 + 2)$ . This simplifies to $u(0.4t + 3.5)$ .
Transform Second Unit Step: Now, let's transform the second unit step function: $u(0.4t + 1.5 - 1)$ . This simplifies to $u(0.4t + 0.5)$ .
Calculate Output Signal: The output signal $y(t)$ is then the difference between these two transformed unit step functions: $y(t) = u(0.4t + 3.5) - u(0.4t + 0.5)$ .

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What kind of transformation converts the graph of

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Question

\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?

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Question

Tess tried to solve the differential equation

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\frac{d y}{d x}=2 x y-6 x

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\quad \frac{d y}{d x}=(y-3) 2 x

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\quad \int \frac{1}{y-3} d y=\int 2 x d x

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\quad|y-3|=e^{x^{2}}+C_{1}

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7

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(A) Tess's work is correct.

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is incorrect. We're not allowed to factor an expression when solving differential equations.

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is incorrect. The right-hand side of the equation should be

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is incorrect. Tess didn't account for adding

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Aditya tried to solve the differential equation

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of the following expressions represents the distance between the sandbag and the target?

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1

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f(x)

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