Inverse Z-transform in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

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easy

Complete the code to import the inverse Z-transform function from sympy.

Signal Processing

from sympy import symbols, [1]

Drag options to blanks, or click blank then click option'

Aztransform

Biztrans

Cztrans

Dinverse_transform

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'ztrans' which is the forward Z-transform.

Using 'ztransform' which is not a valid sympy function.

Using 'inverse_transform' which does not exist.

The function iztrans is used in sympy to compute the inverse Z-transform.

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medium

Complete the code to define the Z-transform variable 'z' as a symbol.

Signal Processing

z = [1]('z')

Drag options to blanks, or click blank then click option'

Asymbols

Bvar

CSymbol

DFunction

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Symbol' which is a class, not a function.

Using 'Function' which defines symbolic functions, not variables.

Using 'var' which is not a sympy function.

The 'symbols' function from sympy is used to define symbolic variables.

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hard

Fix the error in the code to compute the inverse Z-transform of Xz.

Signal Processing

x = [1](Xz, z, n)

Drag options to blanks, or click blank then click option'

Aztrans

Binverse_z

Ciztrans

Dinvztrans

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'ztrans' which computes the forward Z-transform.

Using 'inverse_z' which is not a valid function.

Using 'invztrans' which does not exist in sympy.

The correct function to compute inverse Z-transform in sympy is 'iztrans'.

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hard

Fill both blanks to define the symbolic variable 'n' as an integer and compute inverse Z-transform.

Signal Processing

n = [1]('n', integer=True)
x = iztrans(Xz, z, [2])

Drag options to blanks, or click blank then click option'

Asymbols

BSymbol

Cn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'symbols' which does not accept integer=True.

Passing 'z' instead of 'n' to iztrans.

Using undefined variables.

Use 'Symbol' to define 'n' as an integer variable, and pass 'n' as the time variable in iztrans.

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hard

Fill all three blanks to create a Z-transform expression, define 'n', and compute the inverse Z-transform.

Signal Processing

Xz = 1 / (1 - [1]**-1)
n = [2]('n', integer=True)
x = [3](Xz, z, n)

Drag options to blanks, or click blank then click option'

Az

BSymbol

Ciztrans

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'n' instead of 'z' in the expression.

Using 'symbols' instead of 'Symbol' for 'n'.

Using 'ztrans' instead of 'iztrans' for inverse transform.

The Z-transform expression uses 'z', 'n' is defined as a Symbol, and 'iztrans' computes the inverse Z-transform.