Difference between revisions of "HW4.1 Scott Erdbruegger ECE301Fall2008mboutin" - Rhea

Latest revision as of 09:57, 16 September 2013

x(t)=2cos(\pi/2 t)+3 \,

T=\dfrac{2\pi}{4} \,

T=\pi/2 \,

x(t)=2\dfrac{e^{.5 j t \pi}-e^{-.5 j t \pi}}{2}+3

x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3e^{.5 \pi*0} \,

a-1=1
a0=3
a1=1
a2=0
a3=0
----
[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

@@ Line 1: / Line 1: @@
+[[Category:problem solving]]
+[[Category:ECE301]]
+[[Category:ECE]]
+[[Category:Fourier series]]
+[[Category:signals and systems]]
+== Example of Computation of Fourier series of a CT SIGNAL ==
+A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]]
+----
 ==CT Periodic Signal==
-:<math>x(t)=2cos(\dfrac{\pi}{2}t) \,</math>
+:<math>x(t)=2cos(\pi/2 t)+3 \,</math>
-:<math>T=\dfrac{2\pi}{\pi/2} \,</math>
+:<math>T=\dfrac{2\pi}{4} \,</math>
-:<math>T=4 \,</math>
+:<math>T=\pi/2 \,</math>
-:<math>x(t)=2\dfrac{e^{.5 j t \pi}-e^{-.5 j t \pi}}{2}</math>
+:<math>x(t)=2\dfrac{e^{.5 j t \pi}-e^{-.5 j t \pi}}{2}+3</math>
-:<math>x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi} \, </math>
+:<math>x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3e^{.5 \pi*0} \, </math>
-:<math>ak=\dfrac{1}{T}\int_0^T x(t) e^{-jk\dfrac{2\pi}{T}t} dt </math>
+<pre>
-:<math>a1=\dfrac{1}{4}\int_0^4 x(t) e^{-j\dfrac{2\pi}{4}t} dt </math>
+a-1=1
-:<math>a1=\dfrac{1}{4}\int_0^4 (e^{\dfrac{\pi}{2} j t }-e^{-\dfrac{\pi}{2} j t }) e^{-jt\dfrac{\pi}{2}} dt </math>
+a0=3
-:<math>a1=\dfrac{1}{4}\int_0^4 -e^{-\pi j t } dt </math>
+a1=1
-:<math>a1=\dfrac{1}{4}[\dfrac{e^{-\pi j t }}{\pi}]|_0^4 </math>
+a2=0
+a3=0
+----
+[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]