DTFT (Discrete Time Fourier Transform) duality property applied to cos / sin
$begingroup$
Normal DTFT table contains:
$$
cos(omega_0 n) xrightarrow{DTFT 2pi} pi delta[omega - omega_0] + pi delta[omega + omega_0]
$$
$$
sin(omega_0 n) xrightarrow{DTFT 2pi} i pi delta[omega - omega_0] - i pi delta[omega + omega_0]
$$
How would I apply "duality property" to obtain DTFT transform for:
$$
??? xrightarrow{DTFT 2pi} cos(omega_0 omega)
$$
$$
??? xrightarrow{DTFT 2pi} sin(omega_0 omega)
$$
??? = fill in the blanks.
fourier-transform z-transform
edited Jan 12 at 21:38
user
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
$endgroup$
add a comment |
$begingroup$
Normal DTFT table contains:
$$
cos(omega_0 n) xrightarrow{DTFT 2pi} pi delta[omega - omega_0] + pi delta[omega + omega_0]
$$
$$
sin(omega_0 n) xrightarrow{DTFT 2pi} i pi delta[omega - omega_0] - i pi delta[omega + omega_0]
$$
How would I apply "duality property" to obtain DTFT transform for:
$$
??? xrightarrow{DTFT 2pi} cos(omega_0 omega)
$$
$$
??? xrightarrow{DTFT 2pi} sin(omega_0 omega)
$$
??? = fill in the blanks.
fourier-transform z-transform
edited Jan 12 at 21:38
user
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
$endgroup$
add a comment |
$begingroup$
Normal DTFT table contains:
$$
cos(omega_0 n) xrightarrow{DTFT 2pi} pi delta[omega - omega_0] + pi delta[omega + omega_0]
$$
$$
sin(omega_0 n) xrightarrow{DTFT 2pi} i pi delta[omega - omega_0] - i pi delta[omega + omega_0]
$$
How would I apply "duality property" to obtain DTFT transform for:
$$
??? xrightarrow{DTFT 2pi} cos(omega_0 omega)
$$
$$
??? xrightarrow{DTFT 2pi} sin(omega_0 omega)
$$
??? = fill in the blanks.
fourier-transform z-transform
edited Jan 12 at 21:38
user
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
$endgroup$
Normal DTFT table contains:
$$
cos(omega_0 n) xrightarrow{DTFT 2pi} pi delta[omega - omega_0] + pi delta[omega + omega_0]
$$
$$
sin(omega_0 n) xrightarrow{DTFT 2pi} i pi delta[omega - omega_0] - i pi delta[omega + omega_0]
$$
How would I apply "duality property" to obtain DTFT transform for:
$$
??? xrightarrow{DTFT 2pi} cos(omega_0 omega)
$$
$$
??? xrightarrow{DTFT 2pi} sin(omega_0 omega)
$$
??? = fill in the blanks.
fourier-transform z-transform
fourier-transform z-transform
edited Jan 12 at 21:38
user
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
edited Jan 12 at 21:38
user
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
edited Jan 12 at 21:38
user
5,36211030
edited Jan 12 at 21:38
user
5,36211030
edited Jan 12 at 21:38
user
5,36211030
5,36211030
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
asked Jan 12 at 20:43
MrCasualityMrCasuality
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$
cos(omega) = frac{1}{2} (e^{i omega}+e^{-i omega})
$$
$$
sin(omega) = frac{1}{2i} (e^{i omega}-e^{-i omega})
$$
Then the following DTFT Transforms:
$$
delta[n-d] xrightarrow{DTFT} e^{-i omega d}
$$
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$
cos(omega) = frac{1}{2} (e^{i omega}+e^{-i omega})
$$
$$
sin(omega) = frac{1}{2i} (e^{i omega}-e^{-i omega})
$$
Then the following DTFT Transforms:
$$
delta[n-d] xrightarrow{DTFT} e^{-i omega d}
$$
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
$endgroup$
add a comment |
$begingroup$
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$
cos(omega) = frac{1}{2} (e^{i omega}+e^{-i omega})
$$
$$
sin(omega) = frac{1}{2i} (e^{i omega}-e^{-i omega})
$$
Then the following DTFT Transforms:
$$
delta[n-d] xrightarrow{DTFT} e^{-i omega d}
$$
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
$endgroup$
add a comment |
$begingroup$
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$
cos(omega) = frac{1}{2} (e^{i omega}+e^{-i omega})
$$
$$
sin(omega) = frac{1}{2i} (e^{i omega}-e^{-i omega})
$$
Then the following DTFT Transforms:
$$
delta[n-d] xrightarrow{DTFT} e^{-i omega d}
$$
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
$endgroup$
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$
cos(omega) = frac{1}{2} (e^{i omega}+e^{-i omega})
$$
$$
sin(omega) = frac{1}{2i} (e^{i omega}-e^{-i omega})
$$
Then the following DTFT Transforms:
$$
delta[n-d] xrightarrow{DTFT} e^{-i omega d}
$$
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
answered Jan 12 at 21:16
MrCasualityMrCasuality
82
82
add a comment |
add a comment |
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