12 units right and up 8 units

1. Let x[n] be a signal with x[n] = 0 for n<-1 and n > 3. For each signal given below, determine the values of n for which

valentina_108 [34]

Answer:

a) n<1 and n>5

b) 0 < n < -4

c) n > 2 and n < -2

Step-by-step explanation:

The signal is given by x[n] = 0 for n < -1 and n > 3

The problem asks us to determine the values of n for which it's guaranteed to be zero.

a) x[n-2]

We know that n -2 must be less than -1 or greater than 3.

Therefore we're going to write down our inequalities and solve for n

$n-2$

Therefore for n<1 and n>5 x [n-2] will be zero

b) x [n+ 3]

Similarly, n + 3 must be less than -1 or greater than 3

$n+30$

Therefore for n< -4 and n>0, in other words, for 0 < n < -4 x[n-2] will be zero

c)x [-n + 1]

Similarly, -n+1 must be less than -1 or greater than 3

$-n+13-1\\-n>2\\n$

Therefore, for n > 2 and n < -2 x[-n+1] will be zero

4 0

4 years ago

Graph: f(x) = -3sin (x/2 - π/4)

charle [14.2K]

This sine function has the form y = f(x) = a sin (bx + c), where |a| is the amplitude, b is the frequency, and c determines the phase shift.

The period of this sine curve is 2 pi/b, or (in this problem) 2 pi/(1/2).

The phase shift of this curve is -c/b, or -(pi/4)/(1/2). Simplify this.

The amplitude is |-3|, or just 3.

Draw a set of coordinate axes. Draw light horizontal lines through y=3 and y=-3. These are the max and min. values of this sine function.

Plot the phase shift on your x-axis.

Your sine function will begin at (phase shift value, 0) and continue for one period. Add the period to the phase shift value. Draw a sine curve beginning and ending at these two x-values.

I encourage you to share your calculations and to sketch the graph described here. Then I could give you more specific feedback on your work. Good luck!

3 0

3 years ago

Whats the answer for 8,9,10?<br><br> thanks to the people who answer<3

butalik [34]

Step-by-step explanation:

here is the answer attached