Adding/subtracting sinusoids
$begingroup$
I'm trying to calculate resultant function from adding two sinusoids:
$9sin(omega t + tfracpi3)$ and $-7sin(omega t - tfrac3pi8)$
The correct answer is $14.38sin(omega t + 1.444)$, but I get $14.38sin(omega t + 2.745)$.
My calculations are (first using cosine rule to obtain resultant $v$ as):
$sqrt9^2 + (-7)^2 - (2 cdot 9 cdot (-7) cdot cos(pi - tfracpi3 + tfrac3pi8)) = 14.38$
And the angle (using the sine rule):
$pi - arcsin(|-7| sin(pi - tfracpi3 + tfrac3pi8) / 14.38) = 157$ ° or $2.745$ radians.
continuous-signals
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate resultant function from adding two sinusoids:
$9sin(omega t + tfracpi3)$ and $-7sin(omega t - tfrac3pi8)$
The correct answer is $14.38sin(omega t + 1.444)$, but I get $14.38sin(omega t + 2.745)$.
My calculations are (first using cosine rule to obtain resultant $v$ as):
$sqrt9^2 + (-7)^2 - (2 cdot 9 cdot (-7) cdot cos(pi - tfracpi3 + tfrac3pi8)) = 14.38$
And the angle (using the sine rule):
$pi - arcsin(|-7| sin(pi - tfracpi3 + tfrac3pi8) / 14.38) = 157$ ° or $2.745$ radians.
continuous-signals
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
$endgroup$
2
$begingroup$
Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58
add a comment |
$begingroup$
I'm trying to calculate resultant function from adding two sinusoids:
$9sin(omega t + tfracpi3)$ and $-7sin(omega t - tfrac3pi8)$
The correct answer is $14.38sin(omega t + 1.444)$, but I get $14.38sin(omega t + 2.745)$.
My calculations are (first using cosine rule to obtain resultant $v$ as):
$sqrt9^2 + (-7)^2 - (2 cdot 9 cdot (-7) cdot cos(pi - tfracpi3 + tfrac3pi8)) = 14.38$
And the angle (using the sine rule):
$pi - arcsin(|-7| sin(pi - tfracpi3 + tfrac3pi8) / 14.38) = 157$ ° or $2.745$ radians.
continuous-signals
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
$endgroup$
I'm trying to calculate resultant function from adding two sinusoids:
$9sin(omega t + tfracpi3)$ and $-7sin(omega t - tfrac3pi8)$
The correct answer is $14.38sin(omega t + 1.444)$, but I get $14.38sin(omega t + 2.745)$.
My calculations are (first using cosine rule to obtain resultant $v$ as):
$sqrt9^2 + (-7)^2 - (2 cdot 9 cdot (-7) cdot cos(pi - tfracpi3 + tfrac3pi8)) = 14.38$
And the angle (using the sine rule):
$pi - arcsin(|-7| sin(pi - tfracpi3 + tfrac3pi8) / 14.38) = 157$ ° or $2.745$ radians.
continuous-signals
continuous-signals
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
edited Aug 27 '18 at 18:59
robert bristow-johnson
11k31649
11k31649
asked Aug 27 '18 at 17:59
Bord81Bord81
84
asked Aug 27 '18 at 17:59
Bord81Bord81
84
asked Aug 27 '18 at 17:59
Bord81Bord81
84
84
2
$begingroup$
Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58
add a comment |
2
$begingroup$
Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58
2
2
$begingroup$
Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a trigonometry question, but can also be solved using complex exponentials , which makes it a more DSP type.
We shall use the identitiy:
$$ sin(phi) = frace^jphi - e^-jphi 2j $$
or the more general case:
$$ sin(omega t + phi) = frace^jomega t e^jphi - e^-jomega t e^-jphi 2j $$
and further more general case:
$$
beginalign
|K| sin(omega t + phi + theta_k) &= |K|frace^jomega t e^jphie^jtheta_k - e^-jomega t e^-jphie^-jtheta_k 2j \
&= frace^jomega t e^jphiK - e^-jomega t e^-jphiK^* 2j tag1\
endalign
$$
where $K$ is a complex constant defined as $K = K_r + j K_i = |K| e^jtheta_k $ both in rectangular and polar forms.
Now proceed in decomposing the given signal into complex exponentials:
$$
beginalign
x(t) &= 9 sin(omega t + pi/3) - 7 sin(omega t - 3pi/8) \
&= (9/2j)left( e^jomega t e^jpi/3 - e^-jomega t e^-jpi/3 right) - (7/2j)left( e^jomega t e^-j3pi/8 - e^-jomega t e^j3pi/8 right) \
&= frac e^jomega tleft[9 e^jpi/3 - 7e^-3pi/8 right] - e^-jomega tleft[ 9 e^-jpi/3 - 7e^3pi/8 right] 2j tag2\
&= frac e^jomega tK - e^-jomega tK^* 2j\
endalign
$$
Now denoting $9 e^jpi/3 - 7e^-j3pi/8 = K$, the last line, Eq(2) becomes similar to Eq(1). Now all you need to do is find the magnitude and phase angle of the complex number $K$, which are :
$$ K = 9 e^jpi/3 - 7e^j3pi/8 = 1.8212 + 14.2614 j $$
$$ |K| = 14.3772 $$
$$ theta_k = 1.4438 ~~~text radians $$
Plugging these values gives you the final answer :
$$boxedsin(omega t + theta_k) = 14.38 sin(omega t + 1.4438) $$
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
$endgroup$
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
add a comment |
$begingroup$
The easiest way (to my mind) to solve the problem is to
Use the identity $sin(Apm B) = sin A cos B pm cos A sin B$, substituting the known numerical values of $cos B$ and $sin B$,
Gathering the results to express your sum of sinusoids in the form of $C sin A + D cos A$,
Expressing the resulting function as $sqrtC^2+D^2 sinleft(omega t + thetaright)$
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
$endgroup$
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a trigonometry question, but can also be solved using complex exponentials , which makes it a more DSP type.
We shall use the identitiy:
$$ sin(phi) = frace^jphi - e^-jphi 2j $$
or the more general case:
$$ sin(omega t + phi) = frace^jomega t e^jphi - e^-jomega t e^-jphi 2j $$
and further more general case:
$$
beginalign
|K| sin(omega t + phi + theta_k) &= |K|frace^jomega t e^jphie^jtheta_k - e^-jomega t e^-jphie^-jtheta_k 2j \
&= frace^jomega t e^jphiK - e^-jomega t e^-jphiK^* 2j tag1\
endalign
$$
where $K$ is a complex constant defined as $K = K_r + j K_i = |K| e^jtheta_k $ both in rectangular and polar forms.
Now proceed in decomposing the given signal into complex exponentials:
$$
beginalign
x(t) &= 9 sin(omega t + pi/3) - 7 sin(omega t - 3pi/8) \
&= (9/2j)left( e^jomega t e^jpi/3 - e^-jomega t e^-jpi/3 right) - (7/2j)left( e^jomega t e^-j3pi/8 - e^-jomega t e^j3pi/8 right) \
&= frac e^jomega tleft[9 e^jpi/3 - 7e^-3pi/8 right] - e^-jomega tleft[ 9 e^-jpi/3 - 7e^3pi/8 right] 2j tag2\
&= frac e^jomega tK - e^-jomega tK^* 2j\
endalign
$$
Now denoting $9 e^jpi/3 - 7e^-j3pi/8 = K$, the last line, Eq(2) becomes similar to Eq(1). Now all you need to do is find the magnitude and phase angle of the complex number $K$, which are :
$$ K = 9 e^jpi/3 - 7e^j3pi/8 = 1.8212 + 14.2614 j $$
$$ |K| = 14.3772 $$
$$ theta_k = 1.4438 ~~~text radians $$
Plugging these values gives you the final answer :
$$boxedsin(omega t + theta_k) = 14.38 sin(omega t + 1.4438) $$
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
$endgroup$
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
add a comment |
$begingroup$
This is a trigonometry question, but can also be solved using complex exponentials , which makes it a more DSP type.
We shall use the identitiy:
$$ sin(phi) = frace^jphi - e^-jphi 2j $$
or the more general case:
$$ sin(omega t + phi) = frace^jomega t e^jphi - e^-jomega t e^-jphi 2j $$
and further more general case:
$$
beginalign
|K| sin(omega t + phi + theta_k) &= |K|frace^jomega t e^jphie^jtheta_k - e^-jomega t e^-jphie^-jtheta_k 2j \
&= frace^jomega t e^jphiK - e^-jomega t e^-jphiK^* 2j tag1\
endalign
$$
where $K$ is a complex constant defined as $K = K_r + j K_i = |K| e^jtheta_k $ both in rectangular and polar forms.
Now proceed in decomposing the given signal into complex exponentials:
$$
beginalign
x(t) &= 9 sin(omega t + pi/3) - 7 sin(omega t - 3pi/8) \
&= (9/2j)left( e^jomega t e^jpi/3 - e^-jomega t e^-jpi/3 right) - (7/2j)left( e^jomega t e^-j3pi/8 - e^-jomega t e^j3pi/8 right) \
&= frac e^jomega tleft[9 e^jpi/3 - 7e^-3pi/8 right] - e^-jomega tleft[ 9 e^-jpi/3 - 7e^3pi/8 right] 2j tag2\
&= frac e^jomega tK - e^-jomega tK^* 2j\
endalign
$$
Now denoting $9 e^jpi/3 - 7e^-j3pi/8 = K$, the last line, Eq(2) becomes similar to Eq(1). Now all you need to do is find the magnitude and phase angle of the complex number $K$, which are :
$$ K = 9 e^jpi/3 - 7e^j3pi/8 = 1.8212 + 14.2614 j $$
$$ |K| = 14.3772 $$
$$ theta_k = 1.4438 ~~~text radians $$
Plugging these values gives you the final answer :
$$boxedsin(omega t + theta_k) = 14.38 sin(omega t + 1.4438) $$
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
$endgroup$
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
add a comment |
$begingroup$
This is a trigonometry question, but can also be solved using complex exponentials , which makes it a more DSP type.
We shall use the identitiy:
$$ sin(phi) = frace^jphi - e^-jphi 2j $$
or the more general case:
$$ sin(omega t + phi) = frace^jomega t e^jphi - e^-jomega t e^-jphi 2j $$
and further more general case:
$$
beginalign
|K| sin(omega t + phi + theta_k) &= |K|frace^jomega t e^jphie^jtheta_k - e^-jomega t e^-jphie^-jtheta_k 2j \
&= frace^jomega t e^jphiK - e^-jomega t e^-jphiK^* 2j tag1\
endalign
$$
where $K$ is a complex constant defined as $K = K_r + j K_i = |K| e^jtheta_k $ both in rectangular and polar forms.
Now proceed in decomposing the given signal into complex exponentials:
$$
beginalign
x(t) &= 9 sin(omega t + pi/3) - 7 sin(omega t - 3pi/8) \
&= (9/2j)left( e^jomega t e^jpi/3 - e^-jomega t e^-jpi/3 right) - (7/2j)left( e^jomega t e^-j3pi/8 - e^-jomega t e^j3pi/8 right) \
&= frac e^jomega tleft[9 e^jpi/3 - 7e^-3pi/8 right] - e^-jomega tleft[ 9 e^-jpi/3 - 7e^3pi/8 right] 2j tag2\
&= frac e^jomega tK - e^-jomega tK^* 2j\
endalign
$$
Now denoting $9 e^jpi/3 - 7e^-j3pi/8 = K$, the last line, Eq(2) becomes similar to Eq(1). Now all you need to do is find the magnitude and phase angle of the complex number $K$, which are :
$$ K = 9 e^jpi/3 - 7e^j3pi/8 = 1.8212 + 14.2614 j $$
$$ |K| = 14.3772 $$
$$ theta_k = 1.4438 ~~~text radians $$
Plugging these values gives you the final answer :
$$boxedsin(omega t + theta_k) = 14.38 sin(omega t + 1.4438) $$
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
$endgroup$
This is a trigonometry question, but can also be solved using complex exponentials , which makes it a more DSP type.
We shall use the identitiy:
$$ sin(phi) = frace^jphi - e^-jphi 2j $$
or the more general case:
$$ sin(omega t + phi) = frace^jomega t e^jphi - e^-jomega t e^-jphi 2j $$
and further more general case:
$$
beginalign
|K| sin(omega t + phi + theta_k) &= |K|frace^jomega t e^jphie^jtheta_k - e^-jomega t e^-jphie^-jtheta_k 2j \
&= frace^jomega t e^jphiK - e^-jomega t e^-jphiK^* 2j tag1\
endalign
$$
where $K$ is a complex constant defined as $K = K_r + j K_i = |K| e^jtheta_k $ both in rectangular and polar forms.
Now proceed in decomposing the given signal into complex exponentials:
$$
beginalign
x(t) &= 9 sin(omega t + pi/3) - 7 sin(omega t - 3pi/8) \
&= (9/2j)left( e^jomega t e^jpi/3 - e^-jomega t e^-jpi/3 right) - (7/2j)left( e^jomega t e^-j3pi/8 - e^-jomega t e^j3pi/8 right) \
&= frac e^jomega tleft[9 e^jpi/3 - 7e^-3pi/8 right] - e^-jomega tleft[ 9 e^-jpi/3 - 7e^3pi/8 right] 2j tag2\
&= frac e^jomega tK - e^-jomega tK^* 2j\
endalign
$$
Now denoting $9 e^jpi/3 - 7e^-j3pi/8 = K$, the last line, Eq(2) becomes similar to Eq(1). Now all you need to do is find the magnitude and phase angle of the complex number $K$, which are :
$$ K = 9 e^jpi/3 - 7e^j3pi/8 = 1.8212 + 14.2614 j $$
$$ |K| = 14.3772 $$
$$ theta_k = 1.4438 ~~~text radians $$
Plugging these values gives you the final answer :
$$boxedsin(omega t + theta_k) = 14.38 sin(omega t + 1.4438) $$
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
edited Aug 27 '18 at 21:16
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
answered Aug 27 '18 at 18:57
Fat32Fat32
15.5k31232
15.5k31232
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
add a comment |
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
1
1
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
$begingroup$
yes, thank you for the great explanation!
$endgroup$
– Bord81
Aug 27 '18 at 19:59
1
1
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
Still need some points, but will surely do!)
$endgroup$
– Bord81
Aug 27 '18 at 20:02
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@Fat32 There may be a typo: you give $|K|=14.3772$, but then use $14.438$ in the final equation?
$endgroup$
– MBaz
Aug 27 '18 at 21:00
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
$begingroup$
@MBaz yes there is ! thanks, let me correct.
$endgroup$
– Fat32
Aug 27 '18 at 21:15
add a comment |
$begingroup$
The easiest way (to my mind) to solve the problem is to
Use the identity $sin(Apm B) = sin A cos B pm cos A sin B$, substituting the known numerical values of $cos B$ and $sin B$,
Gathering the results to express your sum of sinusoids in the form of $C sin A + D cos A$,
Expressing the resulting function as $sqrtC^2+D^2 sinleft(omega t + thetaright)$
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
$endgroup$
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
add a comment |
$begingroup$
The easiest way (to my mind) to solve the problem is to
Use the identity $sin(Apm B) = sin A cos B pm cos A sin B$, substituting the known numerical values of $cos B$ and $sin B$,
Gathering the results to express your sum of sinusoids in the form of $C sin A + D cos A$,
Expressing the resulting function as $sqrtC^2+D^2 sinleft(omega t + thetaright)$
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
$endgroup$
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
add a comment |
$begingroup$
The easiest way (to my mind) to solve the problem is to
Use the identity $sin(Apm B) = sin A cos B pm cos A sin B$, substituting the known numerical values of $cos B$ and $sin B$,
Gathering the results to express your sum of sinusoids in the form of $C sin A + D cos A$,
Expressing the resulting function as $sqrtC^2+D^2 sinleft(omega t + thetaright)$
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
$endgroup$
The easiest way (to my mind) to solve the problem is to
Use the identity $sin(Apm B) = sin A cos B pm cos A sin B$, substituting the known numerical values of $cos B$ and $sin B$,
Gathering the results to express your sum of sinusoids in the form of $C sin A + D cos A$,
Expressing the resulting function as $sqrtC^2+D^2 sinleft(omega t + thetaright)$
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
edited Aug 27 '18 at 21:19
Fat32
15.5k31232
15.5k31232
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
answered Aug 27 '18 at 18:28
Dilip SarwateDilip Sarwate
13.2k12463
13.2k12463
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
add a comment |
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
That's more or less the way I'm doing it, but the question is I can't figure out the θ.
$endgroup$
– Bord81
Aug 27 '18 at 18:38
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
$begingroup$
Hint: Set $fracCsqrtC^2+D^2 = cos(theta), fracDsqrtC^2+D^2 = sin(theta) $ and solve $tan(theta)=frac DC$ for $theta$.
$endgroup$
– Dilip Sarwate
Aug 27 '18 at 18:45
add a comment |
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Your question will be much more readable if you format the math properly. I have edited a portion of the question to get you started, and you can find complete instructions here: dsp.stackexchange.com/editing-help#latex. I have also selected a better tag, since your question is unrelated to wavelets.
$endgroup$
– MBaz
Aug 27 '18 at 18:13
$begingroup$
Great! Just be mindful of the "" before "sin" and "cos". I fixed those for you.
$endgroup$
– MBaz
Aug 27 '18 at 18:20
$begingroup$
and lose the asterisks unless you're discussing convolution.
$endgroup$
– robert bristow-johnson
Aug 27 '18 at 18:58