Determine the unit-step response of the discrete-time LTI systems

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SUMMARY

The discussion focuses on determining the unit-step response of discrete-time Linear Time-Invariant (LTI) systems characterized by the impulse response h[n] = (0.9)^{n}e^{j(\pi/2)n}u[n]. The user initially expressed confusion about the process, particularly regarding the summation required for convolution with the unit step function x[n] = u[n]. After consulting with a professor, the user derived the summation formula ∑(0.9)^{k}e^{j(\pi/2)k} from k = 0 to n, utilizing the geometric series property. The user is now seeking clarification on defining the variable 'a' in the context of this summation.

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  • Understanding of discrete-time LTI systems
  • Familiarity with impulse response and unit step functions
  • Knowledge of convolution in signal processing
  • Proficiency in summation techniques and geometric series
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  • Learn about convolution and its application in signal processing
  • Explore geometric series and their convergence criteria
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Students and professionals in electrical engineering, signal processing, or control systems who are working with discrete-time LTI systems and need to understand unit-step responses and convolution techniques.

Mr.Tibbs
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Determine the unit-step response of the discrete-time LTI
systems described by the following impulse responses:

h[n]=(0.9)^{n}e^{j(\pi/2)n}u[n]So I am completely confused. . . I don't even know how to start. . . I want to say that I need to do a summation but the more examples and text I look up the more I'm in the dark. . . any help is appreciated.

The only thing I can think to start is you assume

x[n] = u[n]
 
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How about convolving u[n] with your h[n]?
 
My apologies for taking so long to reply. I was able to talk to my professor and this is what I have now.

\sum(0.9)^{k}e^{j(\pi/2)k}u[n-k] from k = -∞ to ∞.

this turns into

\sum(0.9)^{k}e^{j(\pi/2)k} from k = 0 to n.

using the property of summation :

\sum a^{k} = \frac{1-a^{n+1}}{1-a}

my new snag is what do I define as a?
 

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