- #1
vorcil
- 398
- 0
Take,
[tex] P = 4e^{-j\frac{\pi}{3}} [/tex]
[tex] Q = 4-3j [/tex]
[tex] R = 2e^{j\frac{\pi}{2}} [/tex]
[tex] S = 5 [/tex]
note: I'm using j to be a complex number, it's equivalent to i in mathematics
-
A: express p q r s in both cartesian (a+ib) and polar (re^itheta) forms
B: sketch p q r and s on the complex plane
--------------------------------------------------------------
I'm not too certain as to how P and R work,
I know from the taylor series that r(cos(theta)+isin(theta)) = re^itheta
P is already in polar form, [tex] P = 4e^{-j\frac{\pi}{3}} [/tex]
so the Cartesian form,
4(cos(pi/3)+j(sin(pi/3)))
(a+jb) = [tex] 4 cos\frac{\pi}{3} = a [/tex] and [tex] 4 sin \frac{\pi}{3} = b [/tex]
so it is 2+j(3.464) or [tex] 2+j\sqrt{12} [/tex]
NOTE I AM NOT SURE, BECAUSE THE ORIGINAL EQUATION SHOWS -J
SO SHOULD I PUT, [tex] 2-j\sqrt{12} [/tex] ??
----------------------------------------------------------------
[tex] Q = 4-3j [/tex]
Q is already in cartesian form,
|z| q = 5, = root 4^2 + -(3^2)
[tex] \theta = cos^{-1} \frac{4}{5} = 0.643 rad [/tex]
so the polar form is , 5cis(0.643) = 5(cos(0.643) + jsin(0.643)) = 5e^(j0.643)
----------------------------------------------------------------
[tex] R = 2e^{j\frac{pi}{2}} [/tex]
R is already in polar form,
2 = |z|
2 cos(pi/2) = 0,
2 sin(pi/2) = 2,
this means if i wear to imagine the vector R, it would be going straight up,
so in cartesian form, the equivalent equation of [tex] R = 2e^{j\frac{pi}{2}} [/tex] is j2
-------------------------------------------------------------------------------------------------
S=5
I'm assuming S is already in cartesian form, since it is 5+j0
|z| =5
the angle it makes with the x axis, is 0,
so i think the equation for S in polar form is, [tex] 5e^{j * 0 } [/tex]
[tex] P = 4e^{-j\frac{\pi}{3}} [/tex]
[tex] Q = 4-3j [/tex]
[tex] R = 2e^{j\frac{\pi}{2}} [/tex]
[tex] S = 5 [/tex]
note: I'm using j to be a complex number, it's equivalent to i in mathematics
-
A: express p q r s in both cartesian (a+ib) and polar (re^itheta) forms
B: sketch p q r and s on the complex plane
--------------------------------------------------------------
I'm not too certain as to how P and R work,
I know from the taylor series that r(cos(theta)+isin(theta)) = re^itheta
P is already in polar form, [tex] P = 4e^{-j\frac{\pi}{3}} [/tex]
so the Cartesian form,
4(cos(pi/3)+j(sin(pi/3)))
(a+jb) = [tex] 4 cos\frac{\pi}{3} = a [/tex] and [tex] 4 sin \frac{\pi}{3} = b [/tex]
so it is 2+j(3.464) or [tex] 2+j\sqrt{12} [/tex]
NOTE I AM NOT SURE, BECAUSE THE ORIGINAL EQUATION SHOWS -J
SO SHOULD I PUT, [tex] 2-j\sqrt{12} [/tex] ??
----------------------------------------------------------------
[tex] Q = 4-3j [/tex]
Q is already in cartesian form,
|z| q = 5, = root 4^2 + -(3^2)
[tex] \theta = cos^{-1} \frac{4}{5} = 0.643 rad [/tex]
so the polar form is , 5cis(0.643) = 5(cos(0.643) + jsin(0.643)) = 5e^(j0.643)
----------------------------------------------------------------
[tex] R = 2e^{j\frac{pi}{2}} [/tex]
R is already in polar form,
2 = |z|
2 cos(pi/2) = 0,
2 sin(pi/2) = 2,
this means if i wear to imagine the vector R, it would be going straight up,
so in cartesian form, the equivalent equation of [tex] R = 2e^{j\frac{pi}{2}} [/tex] is j2
-------------------------------------------------------------------------------------------------
S=5
I'm assuming S is already in cartesian form, since it is 5+j0
|z| =5
the angle it makes with the x axis, is 0,
so i think the equation for S in polar form is, [tex] 5e^{j * 0 } [/tex]
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