Ask question

Ask question

All categories

4 years ago

15

Decompose the signal (1 + 0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequenc

y, and phase of each component. Hint: use the identity for cosa cosb.

1 answer:

Alex Ar [27]4 years ago

5 0

Answer:

Given in explanation.

Explanation:

First seperate the signal into components,

$(1 + 0.1 \cos5t)\cos100t=\cos100t+0.1 \cos5t\cos100t$

Now, we will use the identity given in the hint,

$\cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B))\\\\\cos100t+0.1 \cos5t\cos100t=\cos100t + 0.05\cos105t + 0.05\cos95t$

Now the signal is decomposed into its different components.

Accordingly,

$at\:\:100/(2*\pi) = 50/\pi \:Hz \:(16\:Hz)\:\:The \:\:amplitude\:\: is\:\: 1 \\at\:\:105/(2*\pi) = 52.5/\pi \:Hz \:(17\:Hz)\:\:The \:\:amplitude\:\: is\:\: 0.05 \\at\:\:95/(2*\pi) = 47.5/\pi \:Hz \:(15\:Hz)\:\:The \:\:amplitude\:\: is\:\: 0.05$

Since there is no other term within the cos function, the phase differences in all of the above cases are zero.

You might be interested in

A train travels 650 meters in 25 seconds. What is the train's velocity?

frosja888 [35]

The train is traveling 26 meters A second .

3 0

3 years ago

The atomic radii of a divalent cation and a monovalent anion are 0.77 nm and 0.136 nm, respectively.1- Calculate the force of at

MrMuchimi

Answer:

A) attractive force between ions = 5.09x10^-19 N of attractive force

B) there will be no repulsion since both ions have opposite charges.

Explanation:

Detailed explanation and calculation is shown in the image below.

5 0

4 years ago

What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?

Brut [27]

2.31 ft/psig

1.98 psig would equalize feet of head 4’7”

4.583/2.31 = 1.98. .583 is 7”/12”

So round off to 2.0 psig to get steady stream of bubbles

4 0

3 years ago

KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0

4 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago