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If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$.

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$


EDIT: The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $h[n]$ is zero for $n<0$ and $n\ge 7$ (note the "greater or equal" sign). In this case we really have a $6^{th}$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case).

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$.

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$.

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$


EDIT: The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $h[n]$ is zero for $n<0$ and $n\ge 7$ (note the "greater or equal" sign). In this case we really have a $6^{th}$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case).

3 edited body
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If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_1=0.4e^{j\pi/3}$$z_0=0.4e^{j\pi/3}$, $z_1^*$$z_1=z_0^*$, $1/z_1$$z_2=1/z_0$, $1/z_1^*$$z_3=1/z_0^*$, $z_2=3$$z_4=3$, $1/z_2$$z_5=1/z_4$, $z_3=-1$$z_6=-1$. 

If we assume odd symmetry (type IV), the zeros are $z_1=0.4e^{j\pi/3}$$z_0=0.4e^{j\pi/3}$, $z_1^*$$z_1=z_0^*$, $1/z_1$$z_2=1/z_0$, $1/z_1^*$$z_3=1/z_0^*$, $z_2=3$$z_4=3$, $1/z_2$$z_5=1/z_4$, $z_3=1$$z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$.

So if we assume even symmetry (type II), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=-1$. If we assume odd symmetry (type IV), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$. 

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

2 added 495 characters in body
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If the problem description is correct, i.e., if $h[n]$ is non-zerozero for $n=0,\ldots,7$$n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then itthe filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$.

So if we assume even symmetry (type II), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=-1$. If we assume odd symmetry (type IV), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

If the problem description is correct, i.e., if $h[n]$ is non-zero for $n=0,\ldots,7$, then it has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$.

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$.

So if we assume even symmetry (type II), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=-1$. If we assume odd symmetry (type IV), the zeros are $z_1=0.4e^{j\pi/3}$, $z_1^*$, $1/z_1$, $1/z_1^*$, $z_2=3$, $1/z_2$, $z_3=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

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