Suppose you want to find the sum of two sinusoidal voltages, given as follows: v1(t)=V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2)v1(t)=
V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2).
If you stay in the time domain, you will have to use trigonometric identities to perform the addition. But if you transform to the frequency domain, you can simply add the phasors V1V1 and V2V2 as complex numbers using your calculator. Your answer will be a phasor, so you will need to inverse phasor-transform it to get the answer in the time domain. This is an example of a problem that is easier to solve in the frequency domain than in the time domain.
Use phasor techniques to find an expression for v(t) expressed as a single cosine function, where v(t)=[100cos(300t+45∘)+500cos(300t−60∘)] V. Enter your expression using the cosine function. Round real numbers using two digits after the decimal point. Any angles used should be in degrees.
If you stay in the time domain, you will have to use trigonometric identities to perform the addition. But if you transform to the frequency domain, you can simply add the phasors V1V1 and V2V2 as complex numbers using your calculator. Your answer will be a phasor, so you will need to inverse phasor-transform it to get the answer in the time domain. This is an example of a problem that is easier to solve in the frequency domain than in the time domain.
Use phasor techniques to find an expression for v(t) expressed as a single cosine function, where v(t)=[100cos(300t+45∘)+500cos(300t−60∘)] V. Enter your expression using the cosine function. Round real numbers using two digits after the decimal point. Any angles used should be in degrees.
2 answers:
Answer:
Rectangular form V = 320.71 - j362.3
Polar form V = 483.85 < -48.48°
Phasor form V = 483.85cos(300t - 48.48°)
Explanation:
We are given a sinusoidal function
V(t) = 100cos(300t + 45°) + 500cos(300t - 60°)
We are required to find the v(t) expressed as a single cosine function using phasor technique.
In polar form,
100cos(300t + 45°) = 100 < 45°
500cos(300t - 60°) = 500 < -60°
In rectangular form,
100 < 45° = 70.71 + j70.71
500 < -60° = 250 - j433.01
Adding the two signals
(70.71 + j70.71) + (250 - j433.01)
In rectangular form,
V = 320.71 - j362.3
In polar form
V = 483.85 < -48.48°
Therefore, the answer is
in rectangular form V = 320.71 - j362.3
in polar form V = 483.85 < -48.48°
in phasor form V = 483.85cos(300t - 48.48°)
Conversion from Rectangular to Polar form:
V = X + jY to Magnitude < Angle
V = 320.71 - j362.3
Magnitude = = 483.85
Angle = tan⁻¹(Y/X) = tan⁻¹(-362.3/320.71) = -48.48°
V = 483.85 < -48.48°
Conversion from Polar to Rectangular form:
V = 483.85 < -48.48°
X = Magnitude*cos(Angle) and jY = Magnitude*sin(Angle)
X = 483.85*cos(-48.48°) and jY = 483.85*sin(-48.48°)
X = 320.71 and jY = -362.3
V = 320.71 - j362.3
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Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.
Data Given:
x = 27 , 44 , 32 , 47, 23 , 40, 34, 52
y = 30, 19, 24, 13 , 29, 19, 21, 14
It is given that,
∑x = 299
∑y = 167
∑ = 11887
∑ = 3773
We are asked to verify the above values manually in this question.
So,
1. ∑x = 299
Let's verify it:
∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52
∑x = 299
Yes, it is equal to the given value. Hence, verified.
2. ∑y = 167
Let's verify it:
∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14
∑y = 169
No, it is not equal to the given value.
3. ∑ = 11887
Let's verify it:
For this to find, first we need to square all the value of x individually and then add them together to verify.
∑ =
+
+
+
+
+
+
+
∑ = 11,887
Yes, it is equal to the given value. Hence, verified.
4. ∑ = 3773
Let's verify it:
Again, for this we need to find the squares of all the y values and then add them together to verify it.
∑ =
+
+
+
+
+
+
+
∑ = 3,845
No, it is not equal to the given value.
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
